which describes the spacetime curves. Note that we use a minus sign for the square root, since the bodies are radially freely falling toward r=0r=0, i.e., r^(˙) < 0\dot{r}<0.
Usually, when solving equations of this type, one tries to find a solution r=r(tau)r=r(\tau), but one can also find a solution tau=tau(r)\tau=\tau(r), since r(tau)r(\tau) is one-to-one. In principle, one can then find x^(0)x^{0} as
{:(3.1378)x^(0)(tau)=int_(0)^(tau)[1-(r_(**))/(r(tau^(')))]^(-1)Edtau^('):}\begin{equation*}
x^{0}(\tau)=\int_{0}^{\tau}\left[1-\frac{r_{*}}{r\left(\tau^{\prime}\right)}\right]^{-1} E d \tau^{\prime} \tag{3.1378}
\end{equation*}
By using the Euler-Lagrange equations in the same manner as in Problem 2.70, the equations of motion are given by
Since Deltas^(2) >= 0\Delta s^{2} \geq 0 for any physically viable worldline, we obtain r_(0) >= 3r_(**)//2r_{0} \geq 3 r_{*} / 2. Thus, there are no circular orbits inside the r=3r_(**)//2r=3 r_{*} / 2.
2.75
Geodesics for massive test objects in the Schwarzschild metric satisfy the equation of motion
where L=r^(2)varphi^(˙)L=r^{2} \dot{\varphi} is the angular momentum. Approximately circular orbits have an energy close to the minimum of V(r)V(r). To find this minimum, we define U(x)=U(x)=V((1)/(x))V\left(\frac{1}{x}\right) and setU^(')(x_(0))=0\operatorname{set} U^{\prime}\left(x_{0}\right)=0 :
where the solution with a plus sign in front of the square root is unstable. For r_(0)=(1)/(x_(0))≫r_(**)r_{0}=\frac{1}{x_{0}} \gg r_{*}, we require that L^(2)≫r_(**)^(2)L^{2} \gg r_{*}^{2}. The second derivative of the potential is
Thus, finally, the ratio between the periods is (T_(rho))/(T_(varphi))=(1+(9r_(**)^(2))/(4L^(2))+O((r_(**))/(L))^(4))/(1+(3r_(**)^(2))/(2L^(2))+O((r_(**))/(L))^(4))=1+((9)/(4)-(3)/(2))(r_(**)^(2))/(L^(2))+O((r_(**))/(L))^(4)=1+(3r_(**)^(2))/(4L^(2))+O((r_(**))/(L))^(4)\frac{T_{\rho}}{T_{\varphi}}=\frac{1+\frac{9 r_{*}^{2}}{4 L^{2}}+\mathcal{O}\left(\frac{r_{*}}{L}\right)^{4}}{1+\frac{3 r_{*}^{2}}{2 L^{2}}+\mathcal{O}\left(\frac{r_{*}}{L}\right)^{4}}=1+\left(\frac{9}{4}-\frac{3}{2}\right) \frac{r_{*}^{2}}{L^{2}}+\mathcal{O}\left(\frac{r_{*}}{L}\right)^{4}=1+\frac{3 r_{*}^{2}}{4 L^{2}}+\mathcal{O}\left(\frac{r_{*}}{L}\right)^{4}.
For the past-null geodesics to originate at infinity, it is necessary that there exists a classical turning point. The effective potential for a light signal is given by
when using a parametrization such that r^(˙)=1\dot{r}=1 at r rarr oor \rightarrow \infty. For a classical turning point r_(**)r_{*} to exist, it must hold that the maximum of V(r) >= EV(r) \geq E for some rr. The extreme points of VV satisfy (with x=1//rx=1 / r )
For r < oor<\infty, this is solved by r=(3R)/(2)r=\frac{3 R}{2}, where the potential obtains a maximum. We find that the condition for a classical turning point to exist is therefore
Figure 3.13 Illustration of the optical size of a Schwarzschild black hole ("BH"), where bb is the minimal impact parameter of the optical size 4pib^(2)4 \pi b^{2}.
In other words, note that the optical size of the black hole is significantly larger than 4piR^(2)4 \pi R^{2}.
2.77
Using the given ansatz, we have
{:[(3.1405)dZ_(1)=cosh((t)/(2r_(**)))f(r)dt+2r_(**)sinh((t)/(2r_(**)))f^(')(r)dr],[(3.1406)dZ_(2)=sinh((t)/(2r_(**)))f(r)dt+2r_(**)cosh((t)/(2r_(**)))f^(')(r)dr],[(3.1407)dZ_(3)=g^(')(r)dr],[(3.1408)dZ_(4)=sin theta cos varphi dr+r cos theta cos varphi d theta-r sin theta sin varphi d varphi],[(3.1409)dZ_(5)=sin theta sin varphi dr+r cos theta sin varphi d theta+r sin theta cos varphi d varphi],[(3.1410)dZ_(6)=cos theta dr-r sin theta d theta]:}\begin{align*}
d Z_{1} & =\cosh \frac{t}{2 r_{*}} f(r) d t+2 r_{*} \sinh \frac{t}{2 r_{*}} f^{\prime}(r) d r \tag{3.1405}\\
d Z_{2} & =\sinh \frac{t}{2 r_{*}} f(r) d t+2 r_{*} \cosh \frac{t}{2 r_{*}} f^{\prime}(r) d r \tag{3.1406}\\
d Z_{3} & =g^{\prime}(r) d r \tag{3.1407}\\
d Z_{4} & =\sin \theta \cos \varphi d r+r \cos \theta \cos \varphi d \theta-r \sin \theta \sin \varphi d \varphi \tag{3.1408}\\
d Z_{5} & =\sin \theta \sin \varphi d r+r \cos \theta \sin \varphi d \theta+r \sin \theta \cos \varphi d \varphi \tag{3.1409}\\
d Z_{6} & =\cos \theta d r-r \sin \theta d \theta \tag{3.1410}
\end{align*}
Therefore, we find that
{:[(3.1411)dZ_(1)^(2)-dZ_(2)^(2)-dZ_(3)^(2)=f(r)^(2)dt^(2)-[(2r_(**))^(2)f^(')(r)^(2)+g^(')(r)^(2)]dr^(2)],[(3.1412)-dZ_(4)^(2)-dZ_(5)^(2)-dZ_(6)^(2)=-dr^(2)-r^(2)[dtheta^(2)+sin(theta)^(2)dvarphi^(2)]]:}\begin{align*}
d Z_{1}^{2}-d Z_{2}^{2}-d Z_{3}^{2} & =f(r)^{2} d t^{2}-\left[\left(2 r_{*}\right)^{2} f^{\prime}(r)^{2}+g^{\prime}(r)^{2}\right] d r^{2} \tag{3.1411}\\
-d Z_{4}^{2}-d Z_{5}^{2}-d Z_{6}^{2} & =-d r^{2}-r^{2}\left[d \theta^{2}+\sin (\theta)^{2} d \varphi^{2}\right] \tag{3.1412}
\end{align*}
and thus, the result claimed is equivalent to the following conditions
{:(3.1415)g(r)=+-int_(r_(0))^(r)(r^(2)r_(**)+rr_(**)^(2)+r_(**)^(3))/(r^(3))dr:}\begin{equation*}
g(r)= \pm \int_{r_{0}}^{r} \frac{r^{2} r_{*}+r r_{*}^{2}+r_{*}^{3}}{r^{3}} d r \tag{3.1415}
\end{equation*}
with some arbitrary r_(0) > 0r_{0}>0. This proves the claimed result.
2.78
In spherical coordinates, the Minkowski metric is given by
{:(3.1416)ds^(2)=(dx^(0))^(2)-dr^(2)-r^(2)(dtheta^(2)+sin^(2)theta dphi^(2)):}\begin{equation*}
d s^{2}=\left(d x^{0}\right)^{2}-d r^{2}-r^{2}\left(d \theta^{2}+\sin ^{2} \theta d \phi^{2}\right) \tag{3.1416}
\end{equation*}
Using the restriction to the three-dimensional hyperboloid M_(3)M_{3}, i.e., (x^(0))^(2)-r^(2)=\left(x^{0}\right)^{2}-r^{2}=-a^(2)-a^{2}, we find that
{:(3.1417)r=sqrt((x^(0))^(2)+a^(2))=>dr=(x^(0)dx^(0))/(sqrt((x^(0))^(2)+a^(2))):}\begin{equation*}
r=\sqrt{\left(x^{0}\right)^{2}+a^{2}} \Rightarrow d r=\frac{x^{0} d x^{0}}{\sqrt{\left(x^{0}\right)^{2}+a^{2}}} \tag{3.1417}
\end{equation*}
Now, inserting the restriction to M_(3)M_{3} into the Minkowski metric induces the curved metric on M_(3)M_{3}, i.e.,
{:(3.1418)ds^(2)=(a^(2))/((x^(0))^(2)+a^(2))(dx^(0))^(2)-[(x^(0))^(2)+a^(2)](dtheta^(2)+sin^(2)theta dphi^(2)):}\begin{equation*}
d s^{2}=\frac{a^{2}}{\left(x^{0}\right)^{2}+a^{2}}\left(d x^{0}\right)^{2}-\left[\left(x^{0}\right)^{2}+a^{2}\right]\left(d \theta^{2}+\sin ^{2} \theta d \phi^{2}\right) \tag{3.1418}
\end{equation*}
In the case of lightlike geodesics with d phi=0d \phi=0, since phi\phi is assumed to be constant, we have
{:(3.1419)0=ds^(2)=(a^(2))/((x^(0))^(2)+a^(2))(dx^(0))^(2)-[(x^(0))^(2)+a^(2)]dtheta^(2):}\begin{equation*}
0=d s^{2}=\frac{a^{2}}{\left(x^{0}\right)^{2}+a^{2}}\left(d x^{0}\right)^{2}-\left[\left(x^{0}\right)^{2}+a^{2}\right] d \theta^{2} \tag{3.1419}
\end{equation*}
which defines a constant of motion that can be immediately integrated to give theta^(˙)=A//[(x^(0))^(2)+a^(2)]\dot{\theta}=A /\left[\left(x^{0}\right)^{2}+a^{2}\right], where AA is some integration constant (in fact, the constant of motion). Furthermore, the condition ds^(2)=0d s^{2}=0 implies that
where alpha\alpha is an integration constant. Thus, inserting the solution for x^(0)x^{0} into the Euler-Lagrange equation for the theta\theta-coordinate, we obtain the lightlike geodesics with d phi=0d \phi=0 as
where it holds that x^(0)=sqrt(r^(2)-1)x^{0}=\sqrt{r^{2}-1}. The Euler-Lagrange variational equations, i.e., (d)/(ds)((delL)/(delx^(˙)^(mu)))-(delL)/(delx^(mu))=0\frac{d}{d s}\left(\frac{\partial \mathscr{L}}{\partial \dot{x}^{\mu}}\right)-\frac{\partial \mathscr{L}}{\partial x^{\mu}}=0, then become
{:[(3.1426)(d)/(ds)(((r^(˙)))/(r^(2)-1))+rphi^(˙)^(2)+(rr^(˙)^(2))/((r^(2)-1)^(2))=0" for "r],[(3.1427)(d)/(ds)(r^(2)(phi^(˙)))=0" for "phi]:}\begin{align*}
\frac{d}{d s}\left(\frac{\dot{r}}{r^{2}-1}\right)+r \dot{\phi}^{2}+\frac{r \dot{r}^{2}}{\left(r^{2}-1\right)^{2}} & =0 \text { for } r \tag{3.1426}\\
\frac{d}{d s}\left(r^{2} \dot{\phi}\right) & =0 \text { for } \phi \tag{3.1427}
\end{align*}
Furthermore, along a lightlike curve L=0\mathscr{L}=0, and thus, we have
From the Euler-Lagrange variational equation for phi\phi, we find that r^(2)phi^(˙)=A=r^{2} \dot{\phi}=A= const. Inserting this into the lightlike condition, we obtain
so that sqrt(r^(2)-1)=As+s_(0)\sqrt{r^{2}-1}=A s+s_{0} or r=sqrt(1+(As+s_(0))^(2))r=\sqrt{1+\left(A s+s_{0}\right)^{2}}. From r^(2)phi^(˙)=Ar^{2} \dot{\phi}=A, we find that
Now, x^(0)=sqrt(r^(2)-1)=As+s_(0)= tilde(s)=tan(phi-phi_(0))x^{0}=\sqrt{r^{2}-1}=A s+s_{0}=\tilde{s}=\tan \left(\phi-\phi_{0}\right), i.e., x^(0)=tan(phi-phi_(0))x^{0}=\tan \left(\phi-\phi_{0}\right), and finally, Delta phi-=phi-phi_(0)=(pi)/(2)\Delta \phi \equiv \phi-\phi_{0}=\frac{\pi}{2}, which corresponds to Deltax^(0)=tan Delta phi rarr oo\Delta x^{0}=\tan \Delta \phi \rightarrow \infty. Thus, it takes an infinite global time difference Deltax^(0)\Delta x^{0} for a light signal to travel from the point phi_(0)=0\phi_{0}=0 to the point phi=(pi)/(2)\phi=\frac{\pi}{2}. on the surface.
2.80
The Schwarzschild metric (for theta=pi//2\theta=\pi / 2 ) is given by
{:(3.1432)ds^(2)=alpha(dx^(0))^(2)-alpha^(-1)dr^(2)-r^(2)dphi^(2):}\begin{equation*}
d s^{2}=\alpha\left(d x^{0}\right)^{2}-\alpha^{-1} d r^{2}-r^{2} d \phi^{2} \tag{3.1432}
\end{equation*}
where alpha-=alpha(r)=1-(2GM)/(r)\alpha \equiv \alpha(r)=1-\frac{2 G M}{r}. The geodesic equations are derived by varying L=g_(mu nu)x^(˙)^(mu)x^(˙)^(nu)\mathcal{L}=g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}, and therefore, using Euler-Lagrange equations for the x^(0)x^{0} and phi\phi coordinates, we obtain
which are the geodesic equations x^(¨)^(lambda)+Gamma_(mu nu)^(lambda)x^(˙)^(mu)x^(˙)^(nu)=0\ddot{x}^{\lambda}+\Gamma_{\mu \nu}^{\lambda} \dot{x}^{\mu} \dot{x}^{\nu}=0 (see also Problem 2.70). In this case, we actually have the simpler geodesic equations, namely
where g_(00)=alpha,g_(rr)=-(1)/(alpha)g_{00}=\alpha, g_{r r}=-\frac{1}{\alpha}, and g_(phi phi)=-r^(2)g_{\phi \phi}=-r^{2}. Now, the geodesic equation for x^(0)x^{0} implies that x^(˙)^(0)=(E)/( alpha)\dot{x}^{0}=\frac{E}{\alpha}, where EE is a constant of motion, whereas the geodesic equation for phi\phi implies that phi^(˙)=(h)/(r^(2))\dot{\phi}=\frac{h}{r^{2}}, where hh is another constant of motion. Inserting the two constants of motion into the condition for a lightlike geodesic yields
Thus, using the fact that r^(˙)=(dr)/(ds)=f(r)\dot{r}=\frac{d r}{d s}=f(r) and alpha=1-(2GM)/(r)\alpha=1-\frac{2 G M}{r}, we obtain the answer
{:(3.1440)(dr)/(ds)=f(r)=sqrt(E^(2)-(1-(2GM)/(r))(h^(2))/(r^(2))):}\begin{equation*}
\frac{d r}{d s}=f(r)=\sqrt{E^{2}-\left(1-\frac{2 G M}{r}\right) \frac{h^{2}}{r^{2}}} \tag{3.1440}
\end{equation*}
where EE and hh are constants of motion, which is the sought-after differential equation for r(s)r(s) when restricted to the plane theta=(pi)/(2)\theta=\frac{\pi}{2}.
2.81
Consider the metric
{:(3.1441)ds^(2)=g_(mu nu)dx^(mu)dx^(nu)=c^(2)dt^(2)-S(t)^(2)(dx^(2)+dy^(2)+dz^(2)):}\begin{equation*}
d s^{2}=g_{\mu \nu} d x^{\mu} d x^{\nu}=c^{2} d t^{2}-S(t)^{2}\left(d x^{2}+d y^{2}+d z^{2}\right) \tag{3.1441}
\end{equation*}
where S(t)S(t) is an increasing function of time tt with S(0)=0S(0)=0. To find the geodesic equations of motion, we vary the action
{:(3.1442)S=intLds:}\begin{equation*}
S=\int \mathcal{L} d s \tag{3.1442}
\end{equation*}
Now, for lightlike geodesics, we have ds=0d s=0. Note that we may always rotate a coordinate system such that the motion is directed along one spatial coordinate only (say xx ) and the motion is taking place in the positive direction. Then, inserting ds=0d s=0 and the given condition S(t)=t//t_(0)S(t)=t / t_{0} for t_(0) > 0t_{0}>0 into the metric, we find that
{:(3.1449)0=c^(2)dt^(2)-((t)/(t_(0)))^(2)dx^(2)=>cdt=+(t)/(t_(0))dx quad=>quad dx=(ct_(0))/(t)dt:}\begin{equation*}
0=c^{2} d t^{2}-\left(\frac{t}{t_{0}}\right)^{2} d x^{2} \Rightarrow c d t=+\frac{t}{t_{0}} d x \quad \Rightarrow \quad d x=\frac{c t_{0}}{t} d t \tag{3.1449}
\end{equation*}
where e_(r)e_{r} is the initial direction of motion and kk is a constant.
Finally, for the event (or spacetime point) p=(ct_(0),ct_(0),0,0)p=\left(c t_{0}, c t_{0}, 0,0\right), the points on the future light cone are given by
where k=(ct_(0),0,0)k=\left(c t_{0}, 0,0\right). Therefore, the set of events J^(+)(p)J^{+}(p) that are causally connected to pp are the points inside this light cone, i.e.,
From the setup of the problem (see Figure 3.14), we have (at the equator theta=pi//2\theta=\pi / 2 ):
{:[(3.1454)theta^(˙)=1","varphi^(˙)=0","" for the geodesics, "],[(3.1455)X^(theta)=0","X^(varphi)=delta","" for the separation. "]:}\begin{align*}
\dot{\theta}=1, & \dot{\varphi}=0, & & \text { for the geodesics, } \tag{3.1454}\\
X^{\theta}=0, & X^{\varphi}=\delta, & & \text { for the separation. } \tag{3.1455}
\end{align*}
Figure 3.14 Illustration of the setup of the problem. On the unit sphere S^(2)\mathbb{S}^{2}, two geodesics are separated by a small distance delta\delta and both are orthogonal to the equator theta=pi//2\theta=\pi / 2.
Noting that R^(theta)_(theta theta varphi)=g^(theta theta)R_(theta theta theta varphi)=0R^{\theta}{ }_{\theta \theta \varphi}=g^{\theta \theta} R_{\theta \theta \theta \varphi}=0, it thus follows that A^(theta)=0A^{\theta}=0. For R_(theta theta varphi)^(varphi)R_{\theta \theta \varphi}^{\varphi}, we find that
which implies that R_(theta theta varphi)^(varphi)=-1R_{\theta \theta \varphi}^{\varphi}=-1. We therefore conclude that A^(varphi)=-deltaA^{\varphi}=-\delta.
2.83
a) The trajectories x^(mu)(tau)x^{\mu}(\tau) of freely falling particles or photons are the geodesics of the given metric and can be obtained from the variational principle
{:(3.1458)delta int(1)/(2)s^(˙)^(2)d tau=delta int(1)/(2)[t^(˙)^(2)-e^(2t//R)(x^(˙)^(2)+y^(˙)^(2)+z^(˙)^(2))]d tau=0:}\begin{equation*}
\delta \int \frac{1}{2} \dot{s}^{2} d \tau=\delta \int \frac{1}{2}\left[\dot{t}^{2}-e^{2 t / R}\left(\dot{x}^{2}+\dot{y}^{2}+\dot{z}^{2}\right)\right] d \tau=0 \tag{3.1458}
\end{equation*}
with the dot indicating differentiation with respect to the parameter tau\tau. The EulerLagrange equations for this variational problem are
{:[(3.1459)t^(¨)=-(1)/(R)e^(2t//R)(x^(˙)^(2)+y^(˙)^(2)+z^(˙)^(2))],[(3.1460)(d)/(d tau)(-e^(2t//R)x^(˙)^(i))=0]:}\begin{align*}
\ddot{t} & =-\frac{1}{R} e^{2 t / R}\left(\dot{x}^{2}+\dot{y}^{2}+\dot{z}^{2}\right) \tag{3.1459}\\
\frac{d}{d \tau}\left(-e^{2 t / R} \dot{x}^{i}\right) & =0 \tag{3.1460}
\end{align*}
where i=1,2,3i=1,2,3 and (x^(1),x^(2),x^(3))=(x,y,z)\left(x^{1}, x^{2}, x^{3}\right)=(x, y, z). The latter Euler-Lagrange equations can be integrated and give
{:(3.1461)x^(˙)^(i)=c^(i)e^(-2t//R):}\begin{equation*}
\dot{x}^{i}=c^{i} e^{-2 t / R} \tag{3.1461}
\end{equation*}
for certain integration constants c^(i)c^{i}, which are fixed by the initial conditions. Thus, we find
for all i!=ji \neq j, which can be solved in the following simple manner, e.g.,
{:(3.1463)x=a+c^(1)z//c^(3)","quad y=b+c^(2)z//c^(3):}\begin{equation*}
x=a+c^{1} z / c^{3}, \quad y=b+c^{2} z / c^{3} \tag{3.1463}
\end{equation*}
with further integration constants aa and bb. This proves that all geodesics for the given metric are straight lines.
b) The photon moves on a lightlike trajectory along the xx-axis, i.e., ds=dy=d s=d y=dz=0d z=0. This gives
{:(3.1464)dt=-e^(t//R)dx:}\begin{equation*}
d t=-e^{t / R} d x \tag{3.1464}
\end{equation*}
where we assume the minus sign, since we want t^(˙) > 0\dot{t}>0 and x^(˙) < 0\dot{x}<0 at t=0t=0. This implies
{:(3.1465)int_(0)^(t(x))e^(-t//R)dt=-int_(X)^(0)dx:}\begin{equation*}
\int_{0}^{t(x)} e^{-t / R} d t=-\int_{X}^{0} d x \tag{3.1465}
\end{equation*}
a) A suitable coordinate system for the problem is given by using spherical coordinates in the subspace spanned by the spatial directions of the five-dimensional Minkowski space, i.e.,
In these coordinates, dS_(4)\mathrm{dS}_{4} is the restriction of the five-dimensional Minkowski metric to the hyperboloid t^(2)-r^(2)=-T_(0)^(2)t^{2}-r^{2}=-T_{0}^{2}. This condition can be parametrized as
In the coordinates (tau,chi,theta,phi)(\tau, \chi, \theta, \phi), the line element becomes
{:(3.1473)ds^(2)=dtau^(2)-T_(0)^(2)cosh^(2)(tau//T_(0))dOmega^(2):}\begin{equation*}
d s^{2}=d \tau^{2}-T_{0}^{2} \cosh ^{2}\left(\tau / T_{0}\right) d \Omega^{2} \tag{3.1473}
\end{equation*}
where dOmega^(2)=dchi^(2)+sin^(2)(chi)[dtheta^(2)+sin^(2)(theta)dphi^(2)]d \Omega^{2}=d \chi^{2}+\sin ^{2}(\chi)\left[d \theta^{2}+\sin ^{2}(\theta) d \phi^{2}\right] is the standard line element on the three-dimensional sphere.
b) The geodesic equations are given by varying the integral S=intLdsS=\int \mathcal{L} d s, where
This gives T_(mu nu)=-Lambdag_(mu nu)//(8pi G)T_{\mu \nu}=-\Lambda g_{\mu \nu} /(8 \pi G), which is typical for so-called dark energy ( Lambda\Lambda is also know as cosmological constant, in dS_(4)\mathrm{dS}_{4} we have a positive cosmological constant Lambda=3//T_(0)^(2)\Lambda=3 / T_{0}^{2} ).
see Problem 2.27. The trajectory of a light ray then obeys L=0\mathcal{L}=0, i.e., (r^(2)-1)^(2)t^(˙)^(2)=\left(r^{2}-1\right)^{2} \dot{t}^{2}=r^(˙)^(2)\dot{r}^{2}. Taking the square root, we can solve this equation by separation and find that
{:(3.1483)(dr)/(r^(2)-1)=(1)/(2)((1)/(r-1)-(1)/(r+1))=+-dt:}\begin{equation*}
\frac{d r}{r^{2}-1}=\frac{1}{2}\left(\frac{1}{r-1}-\frac{1}{r+1}\right)= \pm d t \tag{3.1483}
\end{equation*}
Integrating, we obtain
{:(3.1484)(1)/(2)(ln((r-1)/(r_(0)-1))-ln((r+1)/(r_(0)+1)))=+-t:}\begin{equation*}
\frac{1}{2}\left(\ln \frac{r-1}{r_{0}-1}-\ln \frac{r+1}{r_{0}+1}\right)= \pm t \tag{3.1484}
\end{equation*}
where we integrated the left-hand side from r_(0)r_{0} to r=r(t)r=r(t) and the right-hand side from t_(0)t_{0} to tt so that r(0)=r_(0)r(0)=r_{0}. Thus, we can write this as
Note that the different signs +-\pm correspond to the two possible directions of the light ray.
b) The trajectory of a particle can also be obtained from the Lagrangian L\mathcal{L}. Using delL//del t=0\partial \mathcal{L} / \partial t=0, we find a conservation law
for some constant c_(1)c_{1}. To compute c_(1)c_{1}, we set t=0t=0 for which it holds that r=r_(0)r=r_{0} and i=gamma=1i=\gamma=1, since the velocity is zero. Thus, we find c_(1)=r_(0)^(2)-1c_{1}=r_{0}^{2}-1, and therefore, we have
A second conservation law is L=c_(2)\mathcal{L}=c_{2} for some other constant c_(2)c_{2}. Inserting again t=0t=0, we find c_(2)=(r_(0)^(2)-1)//2c_{2}=\left(r_{0}^{2}-1\right) / 2, since t^(˙)=1,r^(˙)=0\dot{t}=1, \dot{r}=0, and r=r_(0)r=r_{0} at t=0t=0. Combining the first and the second conservation laws and multiplying with a factor of two, we obtain (r^(2)-1)t^(˙)^(2)-(1)/(r^(2)-1)r^(˙)^(2)=((r_(0)^(2)-1)^(2)-r^(˙)^(2))/(r^(2)-1)=r_(0)^(2)-1quad=>quadr^(˙)^(2)=(r_(0)^(2)-1)(r_(0)^(2)-r^(2))\left(r^{2}-1\right) \dot{t}^{2}-\frac{1}{r^{2}-1} \dot{r}^{2}=\frac{\left(r_{0}^{2}-1\right)^{2}-\dot{r}^{2}}{r^{2}-1}=r_{0}^{2}-1 \quad \Rightarrow \quad \dot{r}^{2}=\left(r_{0}^{2}-1\right)\left(r_{0}^{2}-r^{2}\right).
Separating this equation, we obtain
{:(3.1491)d tau=-(dr)/(sqrt((r_(0)^(2)-1)(r_(0)^(2)-r^(2)))):}\begin{equation*}
d \tau=-\frac{d r}{\sqrt{\left(r_{0}^{2}-1\right)\left(r_{0}^{2}-r^{2}\right)}} \tag{3.1491}
\end{equation*}
where we choose the sign of the square root so that r^(˙) < 0\dot{r}<0. Integrating the left-hand side from 0 to tau\tau and the right-hand side from r_(0)r_{0} to r_(1)r_{1} yields the proper time
When r_(1)rarr1r_{1} \rightarrow 1, the integral diverges as t∼-ln(r_(1)-1)//2t \sim-\ln \left(r_{1}-1\right) / 2. The point r=1r=1 is a singularity similar to the singularity of the Schwarzschild metric at the event horizon: Our computation above shows that the proper time for a particle falling toward r=1r=1 is finite, whereas the coordinate time for that is infinite. Thus, this suggests that r=1r=1 is a coordinate singularity.
2.86
a) Using the given coordinate transformations, we have
{:[(3.1495)dU=cos(alpha)dt-t sin(alpha)d alpha],[(3.1496)dV=sin(alpha)dt+t cos(alpha)d alpha],[(3.1497)dX=sin(theta)cos(varphi)dr+r cos(theta)cos(varphi)d theta-r sin(theta)sin(varphi)d varphi],[(3.1498)dY=sin(theta)sin(varphi)dr+r cos(theta)sin(varphi)d theta+r sin(theta)cos(varphi)d varphi],[(3.1499)dZ=cos(theta)dr-r sin(theta)d theta]:}\begin{align*}
d U & =\cos (\alpha) d t-t \sin (\alpha) d \alpha \tag{3.1495}\\
d V & =\sin (\alpha) d t+t \cos (\alpha) d \alpha \tag{3.1496}\\
d X & =\sin (\theta) \cos (\varphi) d r+r \cos (\theta) \cos (\varphi) d \theta-r \sin (\theta) \sin (\varphi) d \varphi \tag{3.1497}\\
d Y & =\sin (\theta) \sin (\varphi) d r+r \cos (\theta) \sin (\varphi) d \theta+r \sin (\theta) \cos (\varphi) d \varphi \tag{3.1498}\\
d Z & =\cos (\theta) d r-r \sin (\theta) d \theta \tag{3.1499}
\end{align*}
Thus, the metric of AdS_(4)\mathrm{AdS}_{4} in the coordinates alpha,lambda,theta\alpha, \lambda, \theta, and varphi\varphi is given by
{:(3.1500)ds^(2)=dt^(2)+t^(2)dalpha^(2)-dr^(2)-r^(2)dOmega^(2)=cosh(lambda)^(2)dalpha^(2)-dlambda^(2)-sinh(lambda)^(2)dOmega^(2):}\begin{equation*}
d s^{2}=d t^{2}+t^{2} d \alpha^{2}-d r^{2}-r^{2} d \Omega^{2}=\cosh (\lambda)^{2} d \alpha^{2}-d \lambda^{2}-\sinh (\lambda)^{2} d \Omega^{2} \tag{3.1500}
\end{equation*}
where dOmega^(2)=dtheta^(2)+sin(theta)^(2)dvarphi^(2)d \Omega^{2}=d \theta^{2}+\sin (\theta)^{2} d \varphi^{2}.
b) To find the trajectories, we can solve (since theta^(˙)=varphi^(˙)=0\dot{\theta}=\dot{\varphi}=0 )
Contracting the first equation X^(¨)^(a)=2lambdaX^(a)\ddot{X}^{a}=2 \lambda X^{a} with X_(a)X_{a} and using the second equation X^(a)X_(a)=1X^{a} X_{a}=1, we obtain
Thus, we can set X^(˙)^(a)X^(˙)_(a)=0\dot{X}^{a} \dot{X}_{a}=0, which implies that X^(¨)^(a)=0\ddot{X}^{a}=0.
2.87
a) We determine the Christoffel symbols by deriving the geodesic equation with L=g_(mu nu)x^(˙)^(mu)x^(˙)^(nu)=t^(˙)^(2)-a(t)^(2)x^(˙)^(2)\mathcal{L}=g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}=\dot{t}^{2}-a(t)^{2} \dot{x}^{2}. We obtain
with a^(')(t)=da(t)//dta^{\prime}(t)=d a(t) / d t. Comparing with the general form of the geodesic equations x^(¨)^(lambda)+Gamma_(mu nu)^(lambda)x^(˙)^(mu)x^(˙)^(nu)=0\ddot{x}^{\lambda}+\Gamma_{\mu \nu}^{\lambda} \dot{x}^{\mu} \dot{x}^{\nu}=0, we can read off the following nonzero Christoffel symbols
Thus, we find the nonzero components of the Riemann curvature tensor
{:(3.1512)R_(txtx)=g_(tt)R_(xtx)^(t)=aa^('')=R_(xtxt)=-R_(txxt)=-R_(xttx):}\begin{equation*}
R_{t x t x}=g_{t t} R_{x t x}^{t}=a a^{\prime \prime}=R_{x t x t}=-R_{t x x t}=-R_{x t t x} \tag{3.1512}
\end{equation*}
where the last identities follow from general symmetry properties of the Riemann curvature tensor. The other components of the Riemann curvature tensor R_(mu v alpha beta)R_{\mu v \alpha \beta} are zero due to general symmetry properties. Thus, we obtain the components of the Ricci tensor R_(mu nu)R_{\mu \nu} as
{:[(3.1513)R_(tt)=g^(alpha beta)R_(t alpha t beta)=g^(xx)R_(txtx)=-(1)/(a^(2))aa^('')=-(a^(''))/(a)],[(3.1514)R_(xx)=g^(alpha beta)R_(x alpha x beta)=g^(tt)R_(xtxt)=1*aa^('')=aa^('')],[(3.1515)R_(tx)=g^(alpha beta)R_(t alpha x beta)=0],[(3.1516)R_(xt)=g^(alpha beta)R_(x alpha t beta)=0]:}\begin{align*}
R_{t t} & =g^{\alpha \beta} R_{t \alpha t \beta}=g^{x x} R_{t x t x}=-\frac{1}{a^{2}} a a^{\prime \prime}=-\frac{a^{\prime \prime}}{a} \tag{3.1513}\\
R_{x x} & =g^{\alpha \beta} R_{x \alpha x \beta}=g^{t t} R_{x t x t}=1 \cdot a a^{\prime \prime}=a a^{\prime \prime} \tag{3.1514}\\
R_{t x} & =g^{\alpha \beta} R_{t \alpha x \beta}=0 \tag{3.1515}\\
R_{x t} & =g^{\alpha \beta} R_{x \alpha t \beta}=0 \tag{3.1516}
\end{align*}
which can be summarized as R_(mu nu)=-g_(mu nu)a^('')//aR_{\mu \nu}=-g_{\mu \nu} a^{\prime \prime} / a.
Note that the Ricci scalar R-=g^(mu nu)R_(mu nu)R \equiv g^{\mu \nu} R_{\mu \nu} is given by
b) We can find the trajectory of a light ray from L=t^(˙)^(2)-a(t)^(2)x^(˙)^(2)=0\mathcal{L}=\dot{t}^{2}-a(t)^{2} \dot{x}^{2}=0, where t^(˙)=dt//d tau,x^(˙)=dx//d tau\dot{t}=d t / d \tau, \dot{x}=d x / d \tau, and tau\tau being the proper time, i.e.,
where we choose the plus sign of the square root, since it should be assumed that dx//dt > 0d x / d t>0 at t=0t=0. Thus, integrating, we obtain the trajectory of the light ray as
Finally, using the general formula for the geodesic equations x^(¨)^(mu)+Gamma_(v lambda)^(mu)x^(˙)^(nu)x^(˙)^(lambda)=0\ddot{x}^{\mu}+\Gamma_{v \lambda}^{\mu} \dot{x}^{\nu} \dot{x}^{\lambda}=0, we identify the nonzero Christoffel symbols as
For k=1k=1 and d Omega=0d \Omega=0, we can write the Robertson-Walker metric as
{:(3.1555)ds^(2)=c^(2)dt^(2)-S(t)^(2)dchi^(2)=[c^(2)t^(˙)^(2)-S(t)^(2)chi^(˙)^(2)]dtau^(2):}\begin{equation*}
d s^{2}=c^{2} d t^{2}-S(t)^{2} d \chi^{2}=\left[c^{2} \dot{t}^{2}-S(t)^{2} \dot{\chi}^{2}\right] d \tau^{2} \tag{3.1555}
\end{equation*}
where chi=arcsin r\chi=\arcsin r and dot is differentiation with respect to the parameter tau\tau. Thus, we have the metric condition c^(2)t^(˙)^(2)-S(t)^(2)chi^(˙)^(2)=beta=c^{2} \dot{t}^{2}-S(t)^{2} \dot{\chi}^{2}=\beta= const. for an affinely parametrized geodesic.
a) The Euler-Lagrange equations yield the geodesic equations for tt and chi\chi, i.e.,
b) For lightlike geodesics, beta=0\beta=0. Thus, inserting beta=0\beta=0 and alpha=S(t)^(2)chi^(˙)\alpha=S(t)^{2} \dot{\chi} into the result of a), we find that
Now, using the subsidiary condition on chi\chi given in the problem and assuming propagation forward in time, we have the positive derivative, i.e.,
{:(3.1561)(d chi)/(dt)=(c)/(S(t))quad=>quad d chi=(c)/(S(t))dt:}\begin{equation*}
\frac{d \chi}{d t}=\frac{c}{S(t)} \quad \Rightarrow \quad d \chi=\frac{c}{S(t)} d t \tag{3.1561}
\end{equation*}
Integrating leads to
{:(3.1562)Delta chi-=chi_(1)-chi_(0)=int d chi=int_(t_(0))^(t_(1))(c)/(S(t))dt=int_(0)^(T)(c)/(S(t+t_(0)))dt:}\begin{equation*}
\Delta \chi \equiv \chi_{1}-\chi_{0}=\int d \chi=\int_{t_{0}}^{t_{1}} \frac{c}{S(t)} d t=\int_{0}^{T} \frac{c}{S\left(t+t_{0}\right)} d t \tag{3.1562}
\end{equation*}
where t_(1)-=t_(0)+Tt_{1} \equiv t_{0}+T. The coordinate transformation between chi\chi and rr is given by chi=arcsin r\chi=\arcsin r, which means that for r=r_(0)=0r=r_{0}=0, we have chi=chi_(0)=0\chi=\chi_{0}=0, and thus, we find that
Therefore, the distance in the rr-coordinate of the Robertson-Walker metric that a light ray (emitted at universal time t=t_(0)t=t_{0} at r=r_(0)=0r=r_{0}=0 ) travels in the universal time interval [t_(0),t_(0)+T]\left[t_{0}, t_{0}+T\right] is given by
{:(3.1564)Delta r(T)=sin(int_(0)^(T)(c)/(S(t+t_(0)))dt):}\begin{equation*}
\Delta r(T)=\sin \left(\int_{0}^{T} \frac{c}{S\left(t+t_{0}\right)} d t\right) \tag{3.1564}
\end{equation*}
2.91
The given standard Schwarzschild metric is
{:(3.1565)ds^(2)=(1-(r_(**))/(r))c^(2)dt^(2)-(1-(r_(**))/(r))^(-1)dr^(2)-r^(2)dOmega^(2):}\begin{equation*}
d s^{2}=\left(1-\frac{r_{*}}{r}\right) c^{2} d t^{2}-\left(1-\frac{r_{*}}{r}\right)^{-1} d r^{2}-r^{2} d \Omega^{2} \tag{3.1565}
\end{equation*}
where ss is the path parameter, r_(**)-=2GM//c^(2)r_{*} \equiv 2 G M / c^{2}, and dOmega^(2)=dtheta^(2)+sin^(2)theta dphi^(2)d \Omega^{2}=d \theta^{2}+\sin ^{2} \theta d \phi^{2}. In addition, the given initial conditions are
Since (delL)/(del t)=0\frac{\partial \mathcal{L}}{\partial t}=0 and (delL)/(del phi)=0\frac{\partial \mathcal{L}}{\partial \phi}=0, we have two constants of motion stemming from the Euler-Lagrange equations for tt and phi\phi, namely
which means that L=0L=0, independent of the value of theta_(0)\theta_{0} that is not known. Therefore, assuming 0 <= r <= r_(0)0 \leq r \leq r_{0} and 0 <= theta <= pi0 \leq \theta \leq \pi, we have that phi^(˙)=0\dot{\phi}=0. Furthermore, the Euler-Lagrange equation for the theta\theta-coordinate yields
Now, since phi^(˙)=0\dot{\phi}=0, we also have (delL)/(del theta)=-r^(2)sin 2thetaphi^(˙)^(2)=-r^(2)sin 2theta*0=0\frac{\partial \mathcal{L}}{\partial \theta}=-r^{2} \sin 2 \theta \dot{\phi}^{2}=-r^{2} \sin 2 \theta \cdot 0=0, which means that we have another constant of motion stemming from the Euler-Lagrange equation for theta\theta (due to the initial conditions), namely
Again, using the initial conditions, we can determine AA as -2r_(0)^(2)*0=A-2 r_{0}^{2} \cdot 0=A, which also means that A=0A=0. Therefore, assuming 0 <= r <= r_(0)0 \leq r \leq r_{0} as before, we have -2r^(2)theta^(˙)=0-2 r^{2} \dot{\theta}=0, which leads to theta^(˙)=0\dot{\theta}=0. Thus, based on the initial conditions, the given standard Schwarzschild metric is reduced to
{:(3.1573)ds^(2)=(1-(r_(**))/(r))c^(2)dt^(2)-(1-(r_(**))/(r))^(-1)dr^(2):}\begin{equation*}
d s^{2}=\left(1-\frac{r_{*}}{r}\right) c^{2} d t^{2}-\left(1-\frac{r_{*}}{r}\right)^{-1} d r^{2} \tag{3.1573}
\end{equation*}
Now, using the constant of motion EE, given by the initial conditions, we find that
{:(3.1574)i^(˙)=(E)/(2(1-(r_(**))/(r))c^(2))=(1-(r_(**))/(r))^(-1)(E)/(2c^(2))=(dt)/(ds)quad=>quad dt=(1-(r_(**))/(r))^(-1)(E)/(2c^(2))ds:}\begin{equation*}
\dot{i}=\frac{E}{2\left(1-\frac{r_{*}}{r}\right) c^{2}}=\left(1-\frac{r_{*}}{r}\right)^{-1} \frac{E}{2 c^{2}}=\frac{d t}{d s} \quad \Rightarrow \quad d t=\left(1-\frac{r_{*}}{r}\right)^{-1} \frac{E}{2 c^{2}} d s \tag{3.1574}
\end{equation*}
Inserting the expression for dtd t into the reduced standard Schwarzschild metric, we obtain
{:[ds^(2)=(1-(r_(**))/(r))c^(2)(1-(r_(**))/(r))^(-2)(E^(2))/(4c^(4))-(1-(r_(**))/(r))^(-1)dr^(2)],[(3.1575)=(1-(r_(**))/(r))^(-1)(E^(2))/(4c^(2))ds^(2)-(1-(r_(**))/(r))^(-1)dr^(2)]:}\begin{align*}
d s^{2} & =\left(1-\frac{r_{*}}{r}\right) c^{2}\left(1-\frac{r_{*}}{r}\right)^{-2} \frac{E^{2}}{4 c^{4}}-\left(1-\frac{r_{*}}{r}\right)^{-1} d r^{2} \\
& =\left(1-\frac{r_{*}}{r}\right)^{-1} \frac{E^{2}}{4 c^{2}} d s^{2}-\left(1-\frac{r_{*}}{r}\right)^{-1} d r^{2} \tag{3.1575}
\end{align*}
which can be written as
{:(3.1576)dr^(2)=[(E^(2))/(4c^(2))-(1-(r_(**))/(r))]ds^(2)=>((dr)/(ds))=(E^(2))/(4c^(2))-(1-(r_(**))/(r)):}\begin{equation*}
d r^{2}=\left[\frac{E^{2}}{4 c^{2}}-\left(1-\frac{r_{*}}{r}\right)\right] d s^{2} \Rightarrow\left(\frac{d r}{d s}\right)=\frac{E^{2}}{4 c^{2}}-\left(1-\frac{r_{*}}{r}\right) \tag{3.1576}
\end{equation*}
Since the spaceship is freely falling from r_(0)r_{0} toward the true singularity at r=0r=0 in a Schwarzschild black hole, we have r^(˙)=(dr)/(ds) < 0\dot{r}=\frac{d r}{d s}<0, which means that
{:(3.1577)(dr)/(ds)=-sqrt((E^(2))/(4c^(2))-(1-(r_(**))/(r)))=(dr)/(cd tau)=>d tau=-(dr)/(csqrt((E^(2))/(4c^(2))-(1-(r_(**))/(r)))):}\begin{equation*}
\frac{d r}{d s}=-\sqrt{\frac{E^{2}}{4 c^{2}}-\left(1-\frac{r_{*}}{r}\right)}=\frac{d r}{c d \tau} \Rightarrow d \tau=-\frac{d r}{c \sqrt{\frac{E^{2}}{4 c^{2}}-\left(1-\frac{r_{*}}{r}\right)}} \tag{3.1577}
\end{equation*}
where tau\tau is the proper time. Integrating the expression for d taud \tau from r=r_(0)r=r_{0} (at tau=0\tau=0 ) to r=0r=0, we find the proper time tau\tau needed to reach the singularity r=0r=0 (when starting from r=r_(0) < r_(**)r=r_{0}<r_{*} ) as
However, note that (see above) the constant of motion EE can be written in terms of the initial conditions as E=2(1-(r_(**))/(r_(0)))c^(2)betaE=2\left(1-\frac{r_{*}}{r_{0}}\right) c^{2} \beta, which means that the proper time tau\tau is given by
{:(3.1579)tau=int_(0)^(r_(0))(dr)/(csqrt((1-(r_(s))/(r_(0)))^(2)c^(2)beta^(2)-(1-(r_(**))/(r))))-=int_(0)^(r_(0))f(r)dr:}\begin{equation*}
\tau=\int_{0}^{r_{0}} \frac{d r}{c \sqrt{\left(1-\frac{r_{s}}{r_{0}}\right)^{2} c^{2} \beta^{2}-\left(1-\frac{r_{*}}{r}\right)}} \equiv \int_{0}^{r_{0}} f(r) d r \tag{3.1579}
\end{equation*}
Thus, we can finally identify the function f(r)f(r) as
Write the geodesic equations in (x^(0),r,theta,phi)\left(x^{0}, r, \theta, \phi\right) coordinates in the plane theta=(pi)/(2)\theta=\frac{\pi}{2}, namely
where alpha=(2GM)/(c^(2))\alpha=\frac{2 G M}{c^{2}}. For a derivation of the geodesic equations, see Problem 2.70. The geodesic equation for x^(0)x^{0} can be integrated at once, which yields
{:(3.1584)x^(˙)^(0)=(k)/(1-(alpha )/(r))","quad" where "k" is a constant. ":}\begin{equation*}
\dot{x}^{0}=\frac{k}{1-\frac{\alpha}{r}}, \quad \text { where } k \text { is a constant. } \tag{3.1584}
\end{equation*}
On a geodesic for a timelike object, it holds that
where epsilon=k^(2)-1\epsilon=k^{2}-1. Note the minus sign (which is due to the fact that rr is decreasing, i.e., the observer is freely falling toward the center of the black hole, and hence, r^(˙) < 0\dot{r}<0 ). So, we have
where beta=(alpha )/(epsilon)\beta=\frac{\alpha}{\epsilon} and y=r+(beta)/(2)y=r+\frac{\beta}{2}. Thus, the (proper) time needed for the interval r_(0) <=r_{0} \leqr <= r_(1)r \leq r_{1} is given by
where r_(0)=r(tau_(0))=10^(10)kmr_{0}=r\left(\tau_{0}\right)=10^{10} \mathrm{~km} and r_(1)=r(tau_(1))=alphar_{1}=r\left(\tau_{1}\right)=\alpha, i.e., the Schwarzschild horizon. Now, use the initial condition to calculate epsilon\epsilon :
{:(3.1590)-v_(0)=c(dr)/(dx^(0))=c((r^(˙)))/(x^(˙)^(0))quad" at "tau=tau_(0):}\begin{equation*}
-v_{0}=c \frac{d r}{d x^{0}}=c \frac{\dot{r}}{\dot{x}^{0}} \quad \text { at } \tau=\tau_{0} \tag{3.1590}
\end{equation*}
which can be computed analytically using the hint, but the result is too lengthy for it to be useful to display. Computing the integral numerically, the result is Delta tau≃\Delta \tau \simeq5.85615*10^(8)s5.85615 \cdot 10^{8} \mathrm{~s}, which corresponds to about 18.6 years. Thus, the proper time needed for the observer to reach the Schwarzschild horizon is about 18.6 years.
However, the quantities alpha,r_(0)\alpha, r_{0}, and v_(0)v_{0} are given, so the constant epsilon=k^(2)-1\epsilon=k^{2}-1 can be uniquely computed from them. Assuming r_(0)≫alphar_{0} \gg \alpha, qualitative estimates of the parameters are as follows
{:[v_(0)≃(c)/(k)sqrtepsilon","quad epsilon=k^(2)-1≃(v_(0)^(2))/(c^(2))*k^(2)","quad k≃(1)/(sqrt(1-v_(0)^(2)//c^(2)))(∼1)","],[(3.1593)epsilon=k^(2)-1≃(v_(0)^(2))/(c^(2)-v_(0)^(2))(∼(v_(0)^(2))/(c^(2)))","quad" and "quad beta≃(2GM)/(v_(0)^(2))(1-(v_(0)^(2))/(c^(2)))(∼(2GM)/(v_(0)^(2)))","]:}\begin{gather*}
v_{0} \simeq \frac{c}{k} \sqrt{\epsilon}, \quad \epsilon=k^{2}-1 \simeq \frac{v_{0}^{2}}{c^{2}} \cdot k^{2}, \quad k \simeq \frac{1}{\sqrt{1-v_{0}^{2} / c^{2}}}(\sim 1), \\
\epsilon=k^{2}-1 \simeq \frac{v_{0}^{2}}{c^{2}-v_{0}^{2}}\left(\sim \frac{v_{0}^{2}}{c^{2}}\right), \quad \text { and } \quad \beta \simeq \frac{2 G M}{v_{0}^{2}}\left(1-\frac{v_{0}^{2}}{c^{2}}\right)\left(\sim \frac{2 G M}{v_{0}^{2}}\right), \tag{3.1593}
\end{gather*}
where limits within parentheses also hold if it is assumed that v_(0)≪cv_{0} \ll c. Therefore, using the estimates of the parameters, we can approximate the integral for the (proper) time interval Delta tau\Delta \tau as
{:[Delta tau≃-(1)/(v_(0))int_(r_(0))^(r_(1))(dr)/(sqrt(1+(2GM)/(v_(0)^(2)r)))≃(r_(0))/(v_(0))(sqrt(1+(2GM)/(v_(0)^(2)r_(0)))-(2GM)/(v_(0)^(2)r_(0))*arsinhsqrt((v_(0)^(2)r_(0))/(2GM)))],[(3.1594)≃4.32267*10^(8)s∼13.7" years "]:}\begin{align*}
\Delta \tau & \simeq-\frac{1}{v_{0}} \int_{r_{0}}^{r_{1}} \frac{d r}{\sqrt{1+\frac{2 G M}{v_{0}^{2} r}}} \simeq \frac{r_{0}}{v_{0}}\left(\sqrt{1+\frac{2 G M}{v_{0}^{2} r_{0}}}-\frac{2 G M}{v_{0}^{2} r_{0}} \cdot \operatorname{arsinh} \sqrt{\frac{v_{0}^{2} r_{0}}{2 G M}}\right) \\
& \simeq 4.32267 \cdot 10^{8} \mathrm{~s} \sim 13.7 \text { years } \tag{3.1594}
\end{align*}
which means that the approximation underestimates (by about 25%25 \% ) the true value of the proper time, but it gives the correct order of magnitude.
2.93
Due to the Schwarzschild metric components not depending on tt, we can conclude that del_(t)\partial_{t} is a Killing vector field. It follows that a freely falling particle with a worldline that is a geodesic has a constant of motion
where r_(**)=2GMr_{*}=2 G M is the Schwarzschild radius. With the 4 -velocity being V=betadel_(t)+omegadel_(varphi)V=\beta \partial_{t}+\omega \partial_{\varphi}, we generally find that
where r_(**)=2GMr_{*}=2 G M is the Schwarzschild radius and the dot means differentiation with respect to proper time tau\tau. There exists a conservation law L=1\mathcal{L}=1 (remember that we set c=1c=1 ) and we have the Euler-Lagrange equation
where r_(0)r_{0} and r_(1)r_{1} are the initial and final positions of the particle, respectively, and the integration constant CC is determined by the initial velocity of the particle.
Furthermore, to compute the coordinate time, we have (again fixing r^(˙) < 0\dot{r}<0 )
Indeed, the integrand of the integral giving Delta tau\Delta \tau remains finite when r rarrr_(**)r \rightarrow r_{*}, actually the limit is -1//C-1 / C, and thus, Delta tau\Delta \tau is finite as r_(1)rarrr_(**)r_{1} \rightarrow r_{*}, which means that it only takes a finite proper time to reach the even horizon at r_(**)r_{*}. However, the integrand of the integral giving Delta t\Delta t diverges as -((r)/(r_(**))-1)^(-1)-\left(\frac{r}{r_{*}}-1\right)^{-1} when r rarrr_(**)r \rightarrow r_{*}, and thus, Delta t\Delta t diverges as
when r_(1)rarrr_(**)r_{1} \rightarrow r_{*}, which means that it takes an infinite amount of coordinate time to reach the event horizon at r_(**)r_{*}.
In order to reach the same conclusions as above, we can choose C=1C=1 and find explicitly that
{:[Delta tau=-int_(r_(0))^(r_(1))(dr)/(sqrt(1-(1-(r_(**))/(r):})))=-int_(r_(0))^(r_(1))sqrt((r)/(r_(**)))dr=(2)/(3sqrt(r_(**)))(r_(0)^(3//2)-r_(1)^(3//2))],[(3.1617) rarr(2)/(3)(r_(0)sqrt({:(r_(0))/(r_(**))-r_(**)) < ooquad" when "r_(1)rarrr_(**),):}],[Delta t=-int_(r_(0))^(r_(1))(dr)/((1-(r_(**))/(r))sqrt(1-(1-(r_(**))/(r))))=-int_(r_(0))^(r_(1))(1)/(1-(r_(**))/(r))sqrt((r)/(r_(**)))dr],[=(2)/(3)[(r_(0)+3r_(**))sqrt((r_(0))/(r_(**)))-(r_(1)+3r_(**))sqrt((r_(1))/(r_(**)))-3r_(**)(artanhsqrt((r_(0))/(r_(**)))-artanhsqrt((r_(1))/(r_(**))))]],[(3.1618) rarr oo" when "r_(1)rarrr_(**)","]:}\begin{align*}
\Delta \tau & =-\int_{r_{0}}^{r_{1}} \frac{d r}{\left.\sqrt{1-\left(1-\frac{r_{*}}{r}\right.}\right)}=-\int_{r_{0}}^{r_{1}} \sqrt{\frac{r}{r_{*}}} d r=\frac{2}{3 \sqrt{r_{*}}}\left(r_{0}^{3 / 2}-r_{1}^{3 / 2}\right) \\
& \rightarrow \frac{2}{3}\left(r_{0} \sqrt{\left.\frac{r_{0}}{r_{*}}-r_{*}\right)<\infty \quad \text { when } r_{1} \rightarrow r_{*},}\right. \tag{3.1617}\\
\Delta t & =-\int_{r_{0}}^{r_{1}} \frac{d r}{\left(1-\frac{r_{*}}{r}\right) \sqrt{1-\left(1-\frac{r_{*}}{r}\right)}}=-\int_{r_{0}}^{r_{1}} \frac{1}{1-\frac{r_{*}}{r}} \sqrt{\frac{r}{r_{*}}} d r \\
& =\frac{2}{3}\left[\left(r_{0}+3 r_{*}\right) \sqrt{\frac{r_{0}}{r_{*}}}-\left(r_{1}+3 r_{*}\right) \sqrt{\frac{r_{1}}{r_{*}}}-3 r_{*}\left(\operatorname{artanh} \sqrt{\frac{r_{0}}{r_{*}}}-\operatorname{artanh} \sqrt{\frac{r_{1}}{r_{*}}}\right)\right] \\
& \rightarrow \infty \text { when } r_{1} \rightarrow r_{*}, \tag{3.1618}
\end{align*}
since lim_(x rarr1)artanh x rarr oo\lim _{x \rightarrow 1} \operatorname{artanh} x \rightarrow \infty. The choice of C=1C=1 corresponds to t^(˙)=(1-(r_(**))/(r))^(-1)rarr\dot{t}=\left(1-\frac{r_{*}}{r}\right)^{-1} \rightarrow 1 as r rarr oor \rightarrow \infty, i.e., the kinetic energy of the particle is chosen exactly such that it would be at rest at r rarr oor \rightarrow \infty.
2.96
Initially, since the observer is moving tangentially, r^(˙)=0\dot{r}=0. Let us consider the constants of motion for the observer due to del_(t)\partial_{t} and del_(varphi)\partial_{\varphi} being Killing vector fields and the observer following a timelike geodesic. Thus, we define
where gamma^(˙)\dot{\gamma} is the observer's 4 -velocity. We use a coordinate system such that theta=pi//2\theta=\pi / 2 and the initial tangent at r=r_(0)r=r_{0} is therefore
The numbers alpha\alpha and beta\beta are determined from the normalization of the 4 -velocity and that the velocity relative to the stationary frame is v_(0)v_{0} given as
there are two possible situations when the observer will not fall into the black hole
When dV//dr <= 0d V / d r \leq 0, the observer will initially start moving outward. The position r=r_(0)r=r_{0} will then always be a turning point meaning that r >= r_(0)r \geq r_{0} for the entire solution. In the case of dV//dr=0d V / d r=0, the observer will move in a circular orbit.
When dV//dr > 0d V / d r>0, the observer will start moving toward the black hole. The requirement not to fall into the black hole is then that there is a turning point at a smaller value of rr, i.e., that V(r)=V(r_(0))V(r)=V\left(r_{0}\right) for some r < r_(0)r<r_{0}. There will then necessarily be another turning point, which is not accessible, at an even smaller rr. The border of this possibility occurs when both of the turning points coincide, i.e., when V(x)-V(x_(0))V(x)-V\left(x_{0}\right) has a double root different from x_(0)x_{0}.
To see when the first case to applies, we compute dV//dxd V / d x with x=1//rx=1 / r. As dx//dr < 0,dV//dr <= 0d x / d r<0, d V / d r \leq 0 is equivalent with dV//dx=(dV//dr)(dr//dx) >= 0d V / d x=(d V / d r)(d r / d x) \geq 0. We find that
This puts a lower limit on the velocity v_(0)v_{0} for which case 1 applies.
Case 2 is a bit more complicated. In order for three turning points to exist, we must have three real solutions to V(x)-V(x_(0))=0V(x)-V\left(x_{0}\right)=0, where the left-hand side is a third degree polynomial. Luckily, we know that x=x_(0)x=x_{0} is a root of the polynomial and can therefore be factored out and we are left with having to have two real roots for a second degree polynomial. After some algebra, we find that the requirement on v_(0)v_{0} for all roots to be real is
However, we also need to ensure that r=r_(0)r=r_{0} is the turning point at the largest rr. This is the case whenever dV//dx < 0d V / d x<0 and the midpoint of the other solutions is smaller than r_(0)r_{0}. The midpoint 1// bar(r)1 / \bar{r} is found to be located at
Thus, in the region r_(0) <= 3r_(**)r_{0} \leq 3 r_{*}, case 1 always applies and for r_(0) > 3r_(**)r_{0}>3 r_{*}, we find that the requirement not to fall into the black hole is given by
(The expressions are equal when r_(0)=3r_(**)r_{0}=3 r_{*} and otherwise the first expression is always smaller.) The solution to the given problem is therefore that the minimal velocity v_(0)v_{0} is given by
where we have used that drd r is negative to omit the positive root. The proper time to reach the singularity is given by integrating this from r_(0)r_{0} to 0 and we find
Note that the integral is applicable even across the Schwarzschild event horizon as its form removes the coordinate singularity that appears there. The integral itself therefore does not suffer from the coordinate singularity and behaves nicely all the way to the singularity at r=0r=0.
2.97
When the engines are turned off, the satellite will follow a geodesic. Assuming theta=pi//2=\theta=\pi / 2= const., the Lagrangian is given by
Normally, these three equations are sufficient, and the third equation is equivalent to the equation of motion that one obtains from varying the Lagrangian with respect to rr, but it happens that there is an exception to this case and that is precisely when r^(˙)=0\dot{r}=0. Varying the Lagrangian with respect to rr and neglecting all time derivatives of rr gives the equation
We make an interesting observation. We have a minimal radius for which we can have a geodesic circular orbit around a star or a black hole, i.e., R=3r_(**)//2R=3 r_{*} / 2. This orbit will be unstable. Stable circular orbits only exist when the radius RR is larger than 3r_(**)//23 r_{*} / 2, i.e., R > 3r_(**)//2R>3 r_{*} / 2. For the case of the Earth, R > R_(0)≫3r_(**)//2R>R_{0} \gg 3 r_{*} / 2.
2.98
The satellite's orbit is at constant radial distance rr from the Earth, which means that dr=0d r=0.
a) Now, the proper time for the satellite is given by
{:(3.1643)dtau^(2)=ds^(2):}\begin{equation*}
d \tau^{2}=d s^{2} \tag{3.1643}
\end{equation*}
Using the Schwarzschild metric with dr=0d r=0 and d theta=0d \theta=0 (since theta=pi//2\theta=\pi / 2 ) for the satellite's orbit around the Earth and that v=rd phi//dtv=r d \phi / d t, we find that
{:[dtau^(2)=(1-(r_(**))/(r))dt^(2)-r^(2)dphi^(2)=(1-(r_(**))/(r))dt^(2)-(r(d phi)/(dt))^(2)dt^(2)],[(3.1644)=[(1-(r_(**))/(r))-v^(2)]dt^(2)]:}\begin{align*}
d \tau^{2}=\left(1-\frac{r_{*}}{r}\right) d t^{2}-r^{2} d \phi^{2} & =\left(1-\frac{r_{*}}{r}\right) d t^{2}-\left(r \frac{d \phi}{d t}\right)^{2} d t^{2} \\
& =\left[\left(1-\frac{r_{*}}{r}\right)-v^{2}\right] d t^{2} \tag{3.1644}
\end{align*}
which implies that
{:(3.1645)d tau=+-sqrt((1-(r_(**))/(r))-v^(2))dt:}\begin{equation*}
d \tau= \pm \sqrt{\left(1-\frac{r_{*}}{r}\right)-v^{2}} d t \tag{3.1645}
\end{equation*}
where the square root is independent of time tt. Thus, using dt//d tau > 0d t / d \tau>0 and integrating from t=0t=0 to t=T=2pi r//vt=T=2 \pi r / v, we obtain the proper time tau\tau for the satellite to complete one orbit around the Earth as
{:(3.1646)tau=int_(0)^(T)sqrt((1-(r_(**))/(r))-v^(2))dt=Tsqrt((1-(r_(**))/(r))-v^(2))=(2pi r)/(v)sqrt((1-(2GM)/(r))-v^(2)):}\begin{equation*}
\tau=\int_{0}^{T} \sqrt{\left(1-\frac{r_{*}}{r}\right)-v^{2}} d t=T \sqrt{\left(1-\frac{r_{*}}{r}\right)-v^{2}}=\frac{2 \pi r}{v} \sqrt{\left(1-\frac{2 G M}{r}\right)-v^{2}} \tag{3.1646}
\end{equation*}
b) The gravitational potential at the satellite is given by
Now, we calculate the ratio between the coordinate time t=Tt=T and the proper time for the satellite tau\tau. Then, we series expand this ratio for small vv and r_(**)//rr_{*} / r. Therefore, using the result in a) for T//tauT / \tau, we find that
as the 4 -velocity U=alphadel_(t)+betadel_(varphi)U=\alpha \partial_{t}+\beta \partial_{\varphi}, with alpha\alpha and beta\beta constant. It follows that the 4-acceleration is
since the satellite is in free fall. Given the Christoffel symbols of the Schwarzschild metric, the only nontrivial component of this equation is the a=ra=r component
b) For the satellite to complete a full lap, varphi\varphi needs to change by 2pi2 \pi. The proper time Delta s\Delta s it takes for the satellite to travel an angle Delta varphi\Delta \varphi is given by
a) The static observer (1) has constant spatial coordinates. It is therefore convenient to use a parametrization of its worldline in terms of its proper time s_(1)s_{1}, which will be proportional to the coordinate time tt
where we have used the expression for Gamma_(tt)^(r)\Gamma_{t t}^{r}, which is the only nonzero Christoffel symbol on the form Gamma_(tt)^(mu)\Gamma_{t t}^{\mu}. The proper acceleration aa is given by -a^(2)=g(A,A)=-a^{2}=g(A, A)=A^(2)A^{2}, and therefore,
where beta\beta and gamma\gamma are constants and s_(2)s_{2} is the proper time of the observer. By definition, this observer has zero 4-acceleration a=0a=0, which can be used to fix the constants. The 4-velocity takes the form V_(2)=betadel_(t)+gammadel_(varphi)V_{2}=\beta \partial_{t}+\gamma \partial_{\varphi} and the 4-acceleration is therefore given by
The entire 4 -acceleration being zero requires that each component is zero separately. However, all of the components except the rr-component are trivially vanishing and we are left with
where the +-\pm just defines the direction of the orbit and we will pick the positive sign in the future. The normalization is again set by the requirement that V^(2)=1V^{2}=1, leading to
In particular, we can see that this expression has the correct limits as rho rarr1\rho \rightarrow 1 when r_(0)rarr oor_{0} \rightarrow \infty and rho rarr oo\rho \rightarrow \infty as r_(0)rarr3r_(**)//2r_{0} \rightarrow 3 r_{*} / 2. The latter result is expected as the circular geodesic at r_(0)=3r_(**)//2r_{0}=3 r_{*} / 2 is lightlike.
c) The relative gamma factor tilde(gamma)=1//sqrt(1-v^(2))\tilde{\gamma}=1 / \sqrt{1-v^{2}}, where vv is the relative speed, is given by the inner product between the two 4 -velocities, i.e.,
in the limit where r_(0)≫r_(**)=2MGr_{0} \gg r_{*}=2 M G.
2.101
a) The proper time tau\tau can be found from
{:(3.1680)dtau^(2)=ds^(2):}\begin{equation*}
d \tau^{2}=d s^{2} \tag{3.1680}
\end{equation*}
Using the Kerr metric with dr=0d r=0 and d theta=0d \theta=0 (since r=Rr=R and theta=pi//2\theta=\pi / 2 ) for the given orbit, we get
{:[dtau^(2)=(1-(r_(**))/(R))dt^(2)+(2ar_(**))/(R)dtd phi-(R^(2)+a^(2)+(a^(2)r_(**))/(R))dphi^(2)],[=[(1-(r_(**))/(R))+(2ar_(**))/(R)(d phi)/(dt)-(R^(2)+a^(2)+(a^(2)r_(**))/(R))((d phi)/(dt))^(2)]dt^(2)],[(3.1681)=[(1-(r_(**))/(R))+(2ar_(**))/(R^(2))v-(1+(a^(2))/(R^(2))+(a^(2)r_(**))/(R^(3)))v^(2)]dt^(2)]:}\begin{align*}
d \tau^{2} & =\left(1-\frac{r_{*}}{R}\right) d t^{2}+\frac{2 a r_{*}}{R} d t d \phi-\left(R^{2}+a^{2}+\frac{a^{2} r_{*}}{R}\right) d \phi^{2} \\
& =\left[\left(1-\frac{r_{*}}{R}\right)+\frac{2 a r_{*}}{R} \frac{d \phi}{d t}-\left(R^{2}+a^{2}+\frac{a^{2} r_{*}}{R}\right)\left(\frac{d \phi}{d t}\right)^{2}\right] d t^{2} \\
& =\left[\left(1-\frac{r_{*}}{R}\right)+\frac{2 a r_{*}}{R^{2}} v-\left(1+\frac{a^{2}}{R^{2}}+\frac{a^{2} r_{*}}{R^{3}}\right) v^{2}\right] d t^{2} \tag{3.1681}
\end{align*}
since rho^(2)=R^(2)\rho^{2}=R^{2} and v=Rd phi//dtv=R d \phi / d t. This implies that
{:(3.1682)d tau=+-sqrt((1-(r_(**))/(R))+(2ar_(**))/(R^(2))v-(1+(a^(2))/(R^(2))+(a^(2)r_(**))/(R^(3)))v^(2))dt:}\begin{equation*}
d \tau= \pm \sqrt{\left(1-\frac{r_{*}}{R}\right)+\frac{2 a r_{*}}{R^{2}} v-\left(1+\frac{a^{2}}{R^{2}}+\frac{a^{2} r_{*}}{R^{3}}\right) v^{2}} d t \tag{3.1682}
\end{equation*}
where the square root is independent of time tt. Thus, using dt//d tau > 0d t / d \tau>0 and integrating from t=t_(0)=0t=t_{0}=0 to t=T=2pi R//vt=T=2 \pi R / v, we obtain the proper time
{:(3.1683)tau=Tsqrt((1-(r_(**))/(R))+(2ar_(**))/(R^(2))v-(1+(a^(2))/(R^(2))+(a^(2)r_(**))/(R^(3)))v^(2)):}\begin{equation*}
\tau=T \sqrt{\left(1-\frac{r_{*}}{R}\right)+\frac{2 a r_{*}}{R^{2}} v-\left(1+\frac{a^{2}}{R^{2}}+\frac{a^{2} r_{*}}{R^{3}}\right) v^{2}} \tag{3.1683}
\end{equation*}
b) Solving the result in a) for T//tauT / \tau and series expanding in vv and r_(**)//Rr_{*} / R up to orders v^(2)v^{2} and r_(**)//Rr_{*} / R (assuming v^(2)∼r_(**)//Rv^{2} \sim r_{*} / R ), we obtain
2.102
a) If we assume a fixed time t=t_(0)t=t_{0}, we observe that the metric is just a flat metric, i.e.,
{:(3.1685)ds^(2)=-a(t_(0))^(2)e^(2Ct_(0))(dx^(2)+dy^(2)+dz^(2)):}\begin{equation*}
d s^{2}=-a\left(t_{0}\right)^{2} e^{2 C t_{0}}\left(d x^{2}+d y^{2}+d z^{2}\right) \tag{3.1685}
\end{equation*}
where x=rho sin theta cos phi,y=rho sin theta sin phix=\rho \sin \theta \cos \phi, y=\rho \sin \theta \sin \phi, and z=rho cos thetaz=\rho \cos \theta are expressed in the spherical coordinates rho,theta\rho, \theta, and phi\phi. Note that a(t_(0))e^(Ct_(0))a\left(t_{0}\right) e^{C t_{0}} is a constant, since t_(0)t_{0} is fixed and CC is a constant.
b) We choose to use the Cartesian coordinates x,yx, y, and zz. In order to find a geodesic in the spacetime, we use the Lagrangian
where dot means differentiation with respect to proper time tau\tau. Now, the metric does not depend on x,yx, y, and zz, so using the Euler-Lagrange equations (d)/(d tau)((delL)/(delx^(˙)^(')))-\frac{d}{d \tau}\left(\frac{\partial \mathcal{L}}{\partial \dot{x}^{\prime}}\right)-(delL)/(delx^(i))=0\frac{\partial \mathcal{L}}{\partial x^{i}}=0, we have three constants of motion, which are given as follows
and imply that k_(1)^(2)+k_(2)^(2)+k_(3)^(2)=a_(0)^(4)//8k_{1}^{2}+k_{2}^{2}+k_{3}^{2}=a_{0}^{4} / 8. Thus, we obtain
{:(3.1694)t^(˙)^(2)-(1)/(a_(0)^(2)e^(2Ct))(a_(0)^(4))/(8)=1quad=>quadt^(˙)^(2)=1+(a_(0)^(2))/(8)e^(-2Ct)quad=>quad(dt)/(d tau)=+-sqrt(1+(a_(0)^(2))/(8)e^(-2Ct)):}\begin{equation*}
\dot{t}^{2}-\frac{1}{a_{0}^{2} e^{2 C t}} \frac{a_{0}^{4}}{8}=1 \quad \Rightarrow \quad \dot{t}^{2}=1+\frac{a_{0}^{2}}{8} e^{-2 C t} \quad \Rightarrow \quad \frac{d t}{d \tau}= \pm \sqrt{1+\frac{a_{0}^{2}}{8} e^{-2 C t}} \tag{3.1694}
\end{equation*}
which can be separated to
{:(3.1695)d tau=+-(dt)/(sqrt(1+(a_(0)^(2))/(8)e^(-2Ct))):}\begin{equation*}
d \tau= \pm \frac{d t}{\sqrt{1+\frac{a_{0}^{2}}{8} e^{-2 C t}}} \tag{3.1695}
\end{equation*}
Finally, using t^(˙)=dt//d tau > 0\dot{t}=d t / d \tau>0, we calculate the proper time Delta tau\Delta \tau for the free-falling particle between the coordinate times t=0t=0 and t=t_(1)t=t_{1}
{:(3.1696)Delta tau=int_(0)^(t_(1))(dt)/(sqrt(1+(a_(0)^(2))/(8)e^(-2Ct)))=(1)/(C)(artanh((1)/(sqrt(1+(a_(0)^(2))/(8)*e^(-2Ct_(1)))))-artanh((1)/(sqrt(1+(a_(0)^(2))/(8))))):}\begin{equation*}
\Delta \tau=\int_{0}^{t_{1}} \frac{d t}{\sqrt{1+\frac{a_{0}^{2}}{8} e^{-2 C t}}}=\frac{1}{C}\left(\operatorname{artanh} \frac{1}{\sqrt{1+\frac{a_{0}^{2}}{8} \cdot e^{-2 C t_{1}}}}-\operatorname{artanh} \frac{1}{\sqrt{1+\frac{a_{0}^{2}}{8}}}\right) \tag{3.1696}
\end{equation*}
2.103
a) Let the initial and final times be t=0t=0 and t=T_(F)t=T_{F}, respectively. Assuming that v=a(t)R_(0)(d phi)/(dt)=v=a(t) R_{0} \frac{d \phi}{d t}= const. and using
{:(3.1697)d tau=sqrt(1-a(t)^(2)R_(0)^(2)((d phi)/(dt))^(2))dt=sqrt(1-v^(2))dt:}\begin{equation*}
d \tau=\sqrt{1-a(t)^{2} R_{0}^{2}\left(\frac{d \phi}{d t}\right)^{2}} d t=\sqrt{1-v^{2}} d t \tag{3.1697}
\end{equation*}
we obtain the proper time from t=0t=0 to t=T_(F)t=T_{F} as
{:(3.1698)tau=int_(0)^(T_(F))sqrt(1-v^(2))dt=sqrt(1-v^(2))T_(F):}\begin{equation*}
\tau=\int_{0}^{T_{F}} \sqrt{1-v^{2}} d t=\sqrt{1-v^{2}} T_{F} \tag{3.1698}
\end{equation*}
Now, we need to find T_(F)T_{F}. From the fact that vv is a constant and using v=e^(t)R_(0)(d phi)/(dt)v=e^{t} R_{0} \frac{d \phi}{d t}, we can calculate how phi\phi depends on tt :
{:(3.1699)e^(-t)dt=(R_(0))/(v)d phiquad=>quad-(e^(-T_(F))-1)=(R_(0))/(v)2pi:}\begin{equation*}
e^{-t} d t=\frac{R_{0}}{v} d \phi \quad \Rightarrow \quad-\left(e^{-T_{F}}-1\right)=\frac{R_{0}}{v} 2 \pi \tag{3.1699}
\end{equation*}
and together with the initial conditions x^(˙)(tau=0)=A\dot{x}(\tau=0)=A and t(tau=0)=0t(\tau=0)=0, we obtain c_(1)=a(0)^(2)Ac_{1}=a(0)^{2} A, and thus, in total, we find that
Similarly, using the same procedure, we can conclude that y^(˙)=z^(˙)=0\dot{y}=\dot{z}=0. In principle, we can determine tt as a function of tau\tau from the Lagrangian
The equation x^(˙)=a(0)^(2)A//a(t)^(2)\dot{x}=a(0)^{2} A / a(t)^{2} can now be rearranged as
{:(3.1706)(a(t)^(2))/(a(0)^(2)A)dx=d tau:}\begin{equation*}
\frac{a(t)^{2}}{a(0)^{2} A} d x=d \tau \tag{3.1706}
\end{equation*}
Integrating both sides, we obtain
{:(3.1707)tau=int_(X_(0))^(X_(D))(a(t)^(2))/(a(0)^(2)A)dx:}\begin{equation*}
\tau=\int_{X_{0}}^{X_{D}} \frac{a(t)^{2}}{a(0)^{2} A} d x \tag{3.1707}
\end{equation*}
2.105
a) The Schwarzschild metric is given by
{:(3.1708)ds^(2)=(1-(r_(**))/(r))dt^(2)-(1-(r_(**))/(r))^(-1)dr^(2)-r^(2)dOmega^(2):}\begin{equation*}
d s^{2}=\left(1-\frac{r_{*}}{r}\right) d t^{2}-\left(1-\frac{r_{*}}{r}\right)^{-1} d r^{2}-r^{2} d \Omega^{2} \tag{3.1708}
\end{equation*}
See Figure 3.15 for the setup of observers AA and BB. We will assume motion in the plane theta=(pi)/(2)\theta=\frac{\pi}{2} and so d theta=0d \theta=0 and dOmega^(2)=dvarphi^(2)d \Omega^{2}=d \varphi^{2}. From the symmetries of the spacetime (Killing vector fields del_(t)\partial_{t} and del_(varphi)\partial_{\varphi} ), we have the conserved quantities
where ss is the proper time. Normalizing this such that V_(A)=t^(˙)del_(t)+r^(˙)del_(r)=alpha_(0)del_(t)V_{A}=\dot{t} \partial_{t}+\dot{r} \partial_{r}=\alpha_{0} \partial_{t} satisfies V_(A)^(2)=1V_{A}^{2}=1, we find that
For observer BB, the 4-velocity is purely radial and at r=r_(1)r=r_{1} is given by V_(B0)=alpha_(1)del_(t)V_{B 0}=\alpha_{1} \partial_{t}. Normalization of V_(B0)V_{B 0} leads to
Note that, while beta(r)\beta(r) can be computed from the normalization of V_(B)V_{B}, it is not necessary to find the relative velocity.
b) From the normalization of BB 's 4 -velocity, we find that
This is a separable differential equation, which integrates to
{:(3.1720)tau=-int_(r_(1))^(r_(0))(dr)/(sqrt((r_(**))/(r)-(r_(**))/(r_(1))))=int_(r_(0))^(r_(1))sqrt((rr_(1))/(r_(**)(r_(1)-r)))dr:}\begin{equation*}
\tau=-\int_{r_{1}}^{r_{0}} \frac{d r}{\sqrt{\frac{r_{*}}{r}-\frac{r_{*}}{r_{1}}}}=\int_{r_{0}}^{r_{1}} \sqrt{\frac{r r_{1}}{r_{*}\left(r_{1}-r\right)}} d r \tag{3.1720}
\end{equation*}
which is therefore the proper time for BB between r_(1)r_{1} and r_(0)r_{0}.
2.106
a) For an observer with x=x_(0)x=x_{0}, we have dx=0d x=0, and therefore, we find that
{:(3.1721)ds^(2)=x_(0)^(2)dt^(2)quad=>quad(dt)/(ds)=+(1)/(x_(0))quad=>quad t=(s)/(x_(0))+t_(0):}\begin{equation*}
d s^{2}=x_{0}^{2} d t^{2} \quad \Rightarrow \quad \frac{d t}{d s}=+\frac{1}{x_{0}} \quad \Rightarrow \quad t=\frac{s}{x_{0}}+t_{0} \tag{3.1721}
\end{equation*}
The worldline can therefore be described by (choosing t_(0)=0t_{0}=0 )
where ss is the proper time. The 4-acceleration AA is defined by A=grad_(gamma^(˙))gamma^(˙)=(chi^(¨)^(mu)+:}A=\nabla_{\dot{\gamma}} \dot{\gamma}=\left(\ddot{\chi}^{\mu}+\right.{:Gamma_(nu sigma)^(mu)chi^(˙)^(v)chi^(˙)^(sigma))del_(mu)\left.\Gamma_{\nu \sigma}^{\mu} \dot{\chi}^{v} \dot{\chi}^{\sigma}\right) \partial_{\mu}. For our worldline, we have x^(˙)=0,t^(˙)=(1)/(x_(0))\dot{x}=0, \dot{t}=\frac{1}{x_{0}}, and x^(¨)=t^(¨)=0\ddot{x}=\ddot{t}=0, and therefore, we obtain
Thus, we identify Gamma_(tt)^(t)=0\Gamma_{t t}^{t}=0 and Gamma_(tt)^(x)=-(1)/(2)g^(xx)del_(x)x^(2)=-(1)/(2)(-1)2x=x\Gamma_{t t}^{x}=-\frac{1}{2} g^{x x} \partial_{x} x^{2}=-\frac{1}{2}(-1) 2 x=x. Therefore, for our worldline, we have
The proper acceleration alpha\alpha is given by alpha^(2)=-A^(2)=-g(A,A)=-x_(0)^(-2)g(del_(x),del_(x))=\alpha^{2}=-A^{2}=-g(A, A)=-x_{0}^{-2} g\left(\partial_{x}, \partial_{x}\right)=x_(0)^(-2)x_{0}^{-2}, which implies that
We also know that g(gamma^(˙),gamma^(˙))=x^(2)t^(˙)^(2)-x^(˙)^(2)=(Q^(2))/(x^(2))-x^(˙)^(2)=1g(\dot{\gamma}, \dot{\gamma})=x^{2} \dot{t}^{2}-\dot{x}^{2}=\frac{Q^{2}}{x^{2}}-\dot{x}^{2}=1. This leads to
Since we start with (dx)/(dt)=0\frac{d x}{d t}=0, we know that Q^(2)=x_(0)^(2)rarr Q=+x_(0)Q^{2}=x_{0}^{2} \rightarrow Q=+x_{0}. To compute the proper time ss to reach x=0x=0, we need to compute the integral
{:[s=int ds=int_(0)^(x_(0))(ds)/(dx)dx=int_(0)^(x_(0))(dx)/(sqrt((x_(0)^(2))/(x^(2))-1))],[=int_(0)^(x_(0))(xdx)/(sqrt(x_(0)^(2)-x^(2)))={x=x_(0)sin theta,dx=x_(0)cos theta d theta}],[=x_(0)int_(0)^(arcsin 1)(sin theta)/(sqrt(1-sin^(2)theta))cos theta d theta],[(3.1729)=x_(0)int_(0)^(arcsin 1)sin theta d theta=-x_(0)[cos theta]_(0)^(arcsin 1)=x_(0)]:}\begin{align*}
s & =\int d s=\int_{0}^{x_{0}} \frac{d s}{d x} d x=\int_{0}^{x_{0}} \frac{d x}{\sqrt{\frac{x_{0}^{2}}{x^{2}}-1}} \\
& =\int_{0}^{x_{0}} \frac{x d x}{\sqrt{x_{0}^{2}-x^{2}}}=\left\{x=x_{0} \sin \theta, d x=x_{0} \cos \theta d \theta\right\} \\
& =x_{0} \int_{0}^{\arcsin 1} \frac{\sin \theta}{\sqrt{1-\sin ^{2} \theta}} \cos \theta d \theta \\
& =x_{0} \int_{0}^{\arcsin 1} \sin \theta d \theta=-x_{0}[\cos \theta]_{0}^{\arcsin 1}=x_{0} \tag{3.1729}
\end{align*}
Note that the given coordinates are Rindler coordinates on Minkowski space. We can also easily deduce s=x_(0)s=x_{0} by transforming to Minkowski coordinates.
2.107
Using the first relation uv=(2mu-r)e^((r-2mu)//2mu)u v=(2 \mu-r) e^{(r-2 \mu) / 2 \mu}, where u < 0u<0 and v > 0v>0, we obtain by differentiation and Leibniz' rule udv+vdu=-rdre^((r-2mu)//2mu)+(2mu-r)(1)/(2mu)e^((r-2mu)//2mu)dr=-(1)/(2mu)e^((r-2mu)//2mu)rdru d v+v d u=-r d r e^{(r-2 \mu) / 2 \mu}+(2 \mu-r) \frac{1}{2 \mu} e^{(r-2 \mu) / 2 \mu} d r=-\frac{1}{2 \mu} e^{(r-2 \mu) / 2 \mu} r d r.
Similarly, using the second relation t=2mu ln(-v//u)t=2 \mu \ln (-v / u), we obtain
{:(3.1731)(1)/(2mu)dt=(1)/(v)dv-(1)/(u)du:}\begin{equation*}
\frac{1}{2 \mu} d t=\frac{1}{v} d v-\frac{1}{u} d u \tag{3.1731}
\end{equation*}
which implies that
{:(3.1732)udv-vdu=(1)/(2mu)uvdt=(1)/(2mu)*(2mu-r)e^((r-2mu)//2mu)dt:}\begin{equation*}
u d v-v d u=\frac{1}{2 \mu} u v d t=\frac{1}{2 \mu} \cdot(2 \mu-r) e^{(r-2 \mu) / 2 \mu} d t \tag{3.1732}
\end{equation*}
Solving for dud u and dvd v, we find that
{:[(3.1733)2udv=[(1)/(2mu)*(2mu-r)dt-(1)/(2mu)rdr]e^((r-2mu)//2mu)],[(3.1734)2vdu=[-(1)/(2mu)(2mu-r)dt-(1)/(2mu)rdr]e^((r-2mu)//2mu)]:}\begin{align*}
& 2 u d v=\left[\frac{1}{2 \mu} \cdot(2 \mu-r) d t-\frac{1}{2 \mu} r d r\right] e^{(r-2 \mu) / 2 \mu} \tag{3.1733}\\
& 2 v d u=\left[-\frac{1}{2 \mu}(2 \mu-r) d t-\frac{1}{2 \mu} r d r\right] e^{(r-2 \mu) / 2 \mu} \tag{3.1734}
\end{align*}
Thus, we have
{:[4dudv=(1)/(uv)2udv*2vdu],[(3.1735)=(1)/(2mu-r)[-(1)/(4mu^(2))(2mu-r)^(2)dt^(2)+(1)/(4mu^(2))r^(2)dr^(2)]e^((r-2mu)//2mu)]:}\begin{align*}
4 d u d v & =\frac{1}{u v} 2 u d v \cdot 2 v d u \\
& =\frac{1}{2 \mu-r}\left[-\frac{1}{4 \mu^{2}}(2 \mu-r)^{2} d t^{2}+\frac{1}{4 \mu^{2}} r^{2} d r^{2}\right] e^{(r-2 \mu) / 2 \mu} \tag{3.1735}
\end{align*}
and finally, assuming mu-=GM\mu \equiv G M, we obtain the equivalence between the KruskalSzekeres metric and the standard Schwarzschild metric as
{:[(16mu^(2))/(r)*e^(-(r-2mu)//2mu)dudv=(4mu^(2))/(r(2mu-r))[-(1)/(4mu^(2))(2mu-r)^(2)dt^(2)+(1)/(4mu^(2))r^(2)dr^(2)]],[=-(2mu-r)/(r)dt^(2)+(r)/(2mu-r)dr^(2)],[=(1-(2mu)/(r))dt^(2)-(1-(2mu)/(r))^(-1)dr^(2)],[(3.1736)=(1-(2GM)/(r))dx^(0)-(1-(2GM)/(r))^(-1)dr^(2)]:}\begin{align*}
\frac{16 \mu^{2}}{r} \cdot e^{-(r-2 \mu) / 2 \mu} d u d v & =\frac{4 \mu^{2}}{r(2 \mu-r)}\left[-\frac{1}{4 \mu^{2}}(2 \mu-r)^{2} d t^{2}+\frac{1}{4 \mu^{2}} r^{2} d r^{2}\right] \\
& =-\frac{2 \mu-r}{r} d t^{2}+\frac{r}{2 \mu-r} d r^{2} \\
& =\left(1-\frac{2 \mu}{r}\right) d t^{2}-\left(1-\frac{2 \mu}{r}\right)^{-1} d r^{2} \\
& =\left(1-\frac{2 G M}{r}\right) d x^{0}-\left(1-\frac{2 G M}{r}\right)^{-1} d r^{2} \tag{3.1736}
\end{align*}
2.108
The metric is then given by
{:(3.1737)ds^(2)=(16mu^(2))/(r)e^((2mu-r)//2mu)dudv-r^(2)dOmega^(2):}\begin{equation*}
d s^{2}=\frac{16 \mu^{2}}{r} e^{(2 \mu-r) / 2 \mu} d u d v-r^{2} d \Omega^{2} \tag{3.1737}
\end{equation*}
where mu=GM//c^(2)\mu=G M / c^{2} and rr (as well as the time t=t(u,v)t=t(u, v), see below) is a function of uu and vv. The coordinate rr is defined by the equation
Note that f(x)=xe^(x//a)f(x)=x e^{x / a} is monotonically increasing when x > -ax>-a (and f(x) >f(x)>-a//e)-a / e), and therefore, y=f(x)y=f(x) has a unique solution xx for any y > -a//ey>-a / e. We treat uu as a kind of universal time and a timelike vector is future directed if its
Figure 3.16 The four regions K_(1),K_(2),K_(3)K_{1}, K_{2}, K_{3}, and K_(4)K_{4} of Kruskal-Szekeres coordinates uu and vv.
projection to del_(u)\partial_{u} is positive. The orientation (needed in integration) is defined by the ordering ( u,v,theta,phiu, v, \theta, \phi ) of coordinates. Note that the radial null lines (radial light rays) are given by du=0d u=0 or dv=0d v=0.
The Kruskal-Szekeres spacetime can be divided into four regions (see Figure 3.16): region K_(1)K_{1} consists of points u < 0,v > 0u<0, v>0, region K_(2)K_{2} of points u,v > 0u, v>0, region K_(3)K_{3} of points u,v < 0u, v<0, and finally, region K_(4)K_{4} of points u > 0,v < 0u>0, v<0. The boundaries between these regions are nonsingular points for the metric. The only singularities are at the boundary uv=2mu//eu v=2 \mu / e. The region K_(1)K_{1} is equivalent to the outer region of a Schwarzschild spacetime. This is seen by performing the coordinate transformation (u,v,theta,phi)|->(t,r,theta,phi)(u, v, \theta, \phi) \mapsto(t, r, \theta, \phi), where r=r(u,v)r=r(u, v) as above and the Schwarzschild time is t=2mu ln(-v//u)t=2 \mu \ln (-v / u). With a similar coordinate transformation, the region K_(4)K_{4} is also seen to be equivalent to the outer Schwarzschild solution. The region K_(2)K_{2} is equivalent with the Schwarzschild black hole. This equivalence is obtained through the coordinate transformation (u,v,theta,phi)|->(t,r,theta,phi)(u, v, \theta, \phi) \mapsto(t, r, \theta, \phi), where r=r(u,v)r=r(u, v) is the same as before but now t=2mu ln(v//u)t=2 \mu \ln (v / u). The region K_(3)K_{3} is called a "white hole."
It is easy to construct smooth timelike curves, which go from either K_(1)K_{1} or K_(4)K_{4} to the black hole K_(2)K_{2}. However, we will prove that once an observer falls into the black hole K_(2)K_{2}, there is no way to go back to the "normal" regions K_(1)K_{1} and K_(4)K_{4}. Analogously, everything escapes the "white hole" K_(3)K_{3}.
Let x(t)x(t) be the timelike path of an observer. Then, along the path
since x(t)x(t) is timelike and in K_(2)K_{2} it holds that r(del r)/(del u)=-2mu ve^((2mu-r)//2mu) < 0r \frac{\partial r}{\partial u}=-2 \mu v e^{(2 \mu-r) / 2 \mu}<0 and similarly for the vv-coordinate.
The boundary between K_(2)K_{2} and the normal regions is r=2mur=2 \mu (i.e., u=0u=0 or v=0v=0 ). The function r(x(t))r(x(t)) was seen to be decreasing, and therefore, the path x(t)x(t) can never hit the boundary r=2mur=2 \mu. However, the observer entering K_(2)K_{2} has a deplorable future, since it will eventually hit the true singularity r=0r=0, again using the monotonicity of the function r(x(t))r(x(t)).
Note that there is also another singularity, the outer boundary of K_(3)K_{3}. Nevertheless, this is of no great concern, since it is in the past; no future directed timelike curve can enter that singularity.
2.109
For s=0s=0, we have r=2mu,v=v_(0),u=0r=2 \mu, v=v_{0}, u=0 and the constant of motion is then given by
a) The scalar quantity R_(mu nu alpha beta)R^(mu v alpha beta)R_{\mu \nu \alpha \beta} R^{\mu v \alpha \beta} has a singularity at r=0r=0. From this follows that the singularity at r=0r=0 is a physical singularity and not simply due to a bad choice of coordinates (which is the case for the coordinate singularity at r=r_(**)r=r_{*} ).
b) For a radial light signal, we have d theta=d phi=0d \theta=d \phi=0 and ds^(2)=0d s^{2}=0. It follows that
{:(3.1751)dr=+-(1-(r_(**))/(r))dt:}\begin{equation*}
d r= \pm\left(1-\frac{r_{*}}{r}\right) d t \tag{3.1751}
\end{equation*}
For r > r_(**)r>r_{*}, we can see from the metric that tt is the time-coordinate since g_(tt) > 0g_{t t}>0. The forward light cone is therefore given by
i.e., one side of the cone going radially outward and another radially inward. For r < r_(**)r<r_{*}, we instead have g_(rr) > 0g_{r r}>0, and thus, rr represents the time-coordinate with time increasing in the negative rr-direction. Thus, for this case, we have
i.e., the tt-coordinate (which is now a spatial coordinate) can both decrease or increase with time (r)(r).
c) In Kruskal-Szekeres coordinates, we still have d Omega=0d \Omega=0 and ds^(2)=0d s^{2}=0 for the radial light cone. It follows that
{:(3.1754)dU=+-dV:}\begin{equation*}
d U= \pm d V \tag{3.1754}
\end{equation*}
Hence, the light cones are straight lines with slope 1 in these coordinates.
2.111
a) The equation of motion for a massive test particle mm (at position r\mathbf{r} ) in a gravitational potential Phi(r)=-GM//r\Phi(r)=-G M / r according to Newton's mechanics (i.e., Newton's second law) is given by
{:(3.1755)F(r)=mr^(¨)=-m grad Phi(r)=mGM grad(1)/(r)=-(mGM)/(r^(2))e_(r)",":}\begin{equation*}
\mathbf{F}(\mathbf{r})=m \ddot{\mathbf{r}}=-m \nabla \Phi(r)=m G M \nabla \frac{1}{r}=-\frac{m G M}{r^{2}} \mathbf{e}_{r}, \tag{3.1755}
\end{equation*}
where MM is the total point mass of a spherically symmetric source (located at the origin) giving rise to the gravitational potential Phi(r)\Phi(r) (solving Newton's field equation in differential form, i.e., grad^(2)Phi=4pi G rho\nabla^{2} \Phi=4 \pi G \rho with rho\rho being the mass density function), or without the massive test particle, the equation of motion in differential form is
whereas the equations of motion according to general relativity are given by the geodesic equations x^(¨)^(mu)+Gamma_(nu lambda)^(mu)x^(˙)^(v)x^(˙)^(lambda)=0\ddot{x}^{\mu}+\Gamma_{\nu \lambda}^{\mu} \dot{x}^{v} \dot{x}^{\lambda}=0, or more explicitly,
with g_(mu nu)g_{\mu \nu} being the metric. Thus, the Christoffel symbols are combinations of firstorder derivatives of the metric.
In the Newtonian limit, we wish to derive the equation of motion according to Newton's mechanics from the equations of motion according to general relativity. Thus, we compute the geodesic equations using the metric g_(mu nu)=eta_(mu nu)+g_{\mu \nu}=\eta_{\mu \nu}+h_(mu nu)h_{\mu \nu} in the linear approximation, i.e., we neglect higher-order terms in h_(mu nu)h_{\mu \nu}. Note that (eta_(mu nu))=diag(1,-1,-1,-1)\left(\eta_{\mu \nu}\right)=\operatorname{diag}(1,-1,-1,-1). For small velocities, the time component x^(˙)^(0)(sigma)\dot{x}^{0}(\sigma) of the 4 -velocity is much larger than the spatial components. For this reason, we can approximate the geodesic equations as
In the frame, where the source is at rest, the first equation says that we can choose the time tt as the curve parameter, i.e., x^(0)(sigma)=sigma=ctx^{0}(\sigma)=\sigma=c t, and then the second equation becomes
The right-hand side (after multiplication by the mass mm of the test particle) is the gravitational force of the source on mm, so this equation is just Newton's second law, i.e., mr^(¨)=-m grad Phi(r)m \ddot{\mathbf{r}}=-m \nabla \Phi(r).
b) Tidal forces in Newtonian physics are due to differences in gravitational accelerations in neighboring points. Since gravitational accelerations are proportional to the derivatives of the gravitational potential, the differences in gravitational accelerations are second-order derivatives of the gravitational potential, and therefore proportional to r^(-3)r^{-3}, since the gravitational potential itself is proportional to r^(-1)r^{-1}. In general relativity, tidal forces are related to the so-called geodesic deviation, which describes how nearby geodesics separate or converge. They are proportional to second-order derivatives of the metric, since the relativistic gravitational potential in the weak field limit appears as the perturbations to the metric. Thus, tidal forces in general relativity are proportional to the curvature (which is proportional to second-order derivatives of the metric).
2.112
a) In the Newtonian limit (where |h_(mu nu)|≪1\left|h_{\mu \nu}\right| \ll 1 everywhere in spacetime), we have the metric tensor g_(mu nu)g_{\mu \nu}, the Christoffel symbols Gamma_(mu nu)^(lambda)\Gamma_{\mu \nu}^{\lambda}, the Riemann curvature tensor R^(mu)_(nu lambda rho)R^{\mu}{ }_{\nu \lambda \rho}, the Ricci tensor R_(mu nu)R_{\mu \nu}, the Ricci scalar RR, and the Einstein tensor G_(mu nu)G_{\mu \nu} as
where h_(mu nu)=h_(nu mu)h_{\mu \nu}=h_{\nu \mu} and h-=h_(mu)^(mu)h \equiv h_{\mu}^{\mu}. Note that in the weak field limit the Christoffel symbols are first-order derivatives of h_(mu nu)h_{\mu \nu}, whereas the Riemann curvature tensor, the Ricci tensor, the Ricci scalar, and the Einstein tensor are all second-order derivatives of h_(mu nu)h_{\mu \nu}.
Thus, the solution to this problem, i.e., the Ricci tensor in the linear approximation for a metric g_(mu nu)=eta_(mu nu)+h_(mu nu)g_{\mu \nu}=\eta_{\mu \nu}+h_{\mu \nu}, is given by
b) Consider the coordinate transformation x^(mu)|->x^('mu)=x^(mu)+chi^(mu)x^{\mu} \mapsto x^{\prime \mu}=x^{\mu}+\chi^{\mu}, where |del_(v)chi^(mu)|≪1\left|\partial_{v} \chi^{\mu}\right| \ll 1. Differentiating this coordinate transformation yields
Using this inverse coordinate transformation for the metric tensor and the definition g_(mu nu)^((1))=eta_(mu nu)+h_(mu nu)g_{\mu \nu}^{(1)}=\eta_{\mu \nu}+h_{\mu \nu}, we find that
which is the gauge transformation of h_(mu nu)h_{\mu \nu}.
c) Consider a coordinate system in which the harmonic (or Lorenz) gauge condition holds, i.e., del^(mu) bar(h)_(mu nu)=0\partial^{\mu} \bar{h}_{\mu \nu}=0, where
Note that bar(h)_(mu)^(mu)=-h_(mu)^(mu)(\bar{h}_{\mu}^{\mu}=-h_{\mu}^{\mu}( or bar(h)=-h)\bar{h}=-h). The tensor bar(h)_(mu nu)\bar{h}_{\mu \nu} has the following gauge transformation
where Phi(x)\Phi(\mathbf{x}) is independent of time tt. Using the Euler-Lagrange equations, we find that (d)/(d tau)(delL)/(del(t^(˙)))-(delL)/(del t)=0quad=>quad(d)/(d tau)[(1-2Phi(x))t^(˙)]=0quad=>quad(1-2Phi(x))t^(˙)=\frac{d}{d \tau} \frac{\partial \mathcal{L}}{\partial \dot{t}}-\frac{\partial \mathcal{L}}{\partial t}=0 \quad \Rightarrow \quad \frac{d}{d \tau}[(1-2 \Phi(\mathbf{x})) \dot{t}]=0 \quad \Rightarrow \quad(1-2 \Phi(\mathbf{x})) \dot{t}= const.,
However, since the metric only holds in the weak-field limit (where only lowestorder terms of Phi\Phi are kept), we should write the Euler-Lagrange equations in this limit and have
Furthermore, we apply the nonrelativistic limit, which says that x^(˙)^(i)≪t^(˙)\dot{x}^{i} \ll \dot{t}, since it holds for a slowly moving massive particle that tau~~t\tau \approx t and dx^(i)//dt≪1d x^{i} / d t \ll 1, which implies that x^(˙)^(i)≪t^(˙)~~1\dot{x}^{i} \ll \dot{t} \approx 1. Thus, we obtain the simple equations
which are the geodesic equations in the nonrelativistic and weak-field limits. The first equation means that energy (or t^(˙)\dot{t} ) is conserved (the integration constant is chosen such that t^(˙)=1\dot{t}=1, which then coincides with L=1\mathcal{L}=1 in the nonrelativistic limit), whereas multiplying the second equation with the mass mm of the massive test particle leads to
which is just Newton's second law, i.e., ma=F=-m grad Phi(x)m \mathbf{a}=\mathbf{F}=-m \nabla \Phi(\mathbf{x}), where the force F=-m grad Phi(x)\mathbf{F}=-m \nabla \Phi(\mathbf{x}) is that of a weak gravitational potential Phi(x)\Phi(\mathbf{x}). Compare the solution to Problem 2.111 a).
b) Using the metric for the weak gravitational potential Phi(x)\Phi(\mathbf{x}), we have
Now, assuming Phi(x)=-gz\Phi(\mathbf{x})=-g z, we find that
{:(3.1790)g_(00)(z)=1+(2)/(c^(2))gz:}\begin{equation*}
g_{00}(z)=1+\frac{2}{c^{2}} g z \tag{3.1790}
\end{equation*}
Then, using the formula for the gravitational redshift to compute the photon's gravitational redshift, we obtain
{:[(omega^('))/(omega)=sqrt((1+(2)/(c^(2))g*0)/(1+(2)/(c^(2))g*h))≃1-(gh)/(c^(2))],[(3.1791)quad=>quad z=(lambda^('))/(lambda)-1=(omega)/(omega^('))-1=sqrt(1+(2gh)/(c^(2)))-1≃(gh)/(c^(2))]:}\begin{align*}
& \frac{\omega^{\prime}}{\omega}=\sqrt{\frac{1+\frac{2}{c^{2}} g \cdot 0}{1+\frac{2}{c^{2}} g \cdot h}} \simeq 1-\frac{g h}{c^{2}} \\
& \quad \Rightarrow \quad z=\frac{\lambda^{\prime}}{\lambda}-1=\frac{\omega}{\omega^{\prime}}-1=\sqrt{1+\frac{2 g h}{c^{2}}}-1 \simeq \frac{g h}{c^{2}} \tag{3.1791}
\end{align*}
where h > 0h>0. Thus, a photon emitted at a lower gravitational potential with an angular frequency omega\omega is received at a higher gravitational potential with a smaller angular frequency omega^(')\omega^{\prime}, i.e., it is redshifted, although the emitter and the receiver are not in relative motion.
Finally, consider a particular frame to be at rest when the photon starts its upward climb in the gravitational potential and falls freely after that. Since the photon climbs from z=0z=0 to z=h > 0z=h>0, i.e., a distance hh, it takes the time Delta t=h//c\Delta t=h / c to arrive at the higher gravitational potential. During this time, the frame has obtained the velocity v=g Delta t=gh//cv=g \Delta t=g h / c downward relative to the higher gravitational potential. Therefore, using formula for the Doppler shift, the photon's angular frequency omega_(ff)\omega_{\mathrm{ff}} relative to the freely falling frame is
{:(3.1792)(omega_(ff))/(omega^('))=sqrt((1-(-gh//c^(2)))/(1+(-gh//c^(2))))≃1+(gh)/(c^(2)):}\begin{equation*}
\frac{\omega_{\mathrm{ff}}}{\omega^{\prime}}=\sqrt{\frac{1-\left(-g h / c^{2}\right)}{1+\left(-g h / c^{2}\right)}} \simeq 1+\frac{g h}{c^{2}} \tag{3.1792}
\end{equation*}
Thus, there is no redshift in a freely falling frame, confirming that it is a local inertial frame. This is the essence of Einstein's equivalence principle, which means that the gravitational field can be transformed away by transforming to an appropriate accelerating ("freely falling") frame of reference.
2.114
a) Consider two massive particles moving freely on two close paths, where one particle has the spacetime path x^(mu)(tau)x^{\mu}(\tau) and the other one has the spacetime path x^(mu)(tau)+s^(mu)(tau)x^{\mu}(\tau)+s^{\mu}(\tau). Therefore, the two particles are separated by the displacement vector s^(mu)(tau)s^{\mu}(\tau), which is small. The geodesic equations (or the equations of motion) for the two particles are given by
Since s^(mu)s^{\mu} is small, we can series expand the Christoffel symbols Gamma_(alpha beta)^(mu)(x+s)\Gamma_{\alpha \beta}^{\mu}(x+s) in the second equation as
where we used Gamma_(alpha beta)^(mu)(x)=Gamma_(beta alpha)^(mu)(x)\Gamma_{\alpha \beta}^{\mu}(x)=\Gamma_{\beta \alpha}^{\mu}(x). Using the definition of the parallel transport of a vector V^(mu)V^{\mu} along a path x^(mu)(tau)x^{\mu}(\tau), i.e.,
In order to simplify the second derivative, we use the above-derived expressions for s^(¨)^(mu)\ddot{s}^{\mu} and x^(¨)^(beta)\ddot{x}^{\beta} and obtain
where in the last step we relabeled some of the summation indices. It holds that del_(alpha)Gamma_(lambda beta)^(mu)s^(lambda)x^(˙)^(alpha)x^(˙)^(beta)=del_(beta)Gamma_(lambda alpha)^(mu)s^(lambda)x^(˙)^(alpha)x^(˙)^(beta)\partial_{\alpha} \Gamma_{\lambda \beta}^{\mu} s^{\lambda} \dot{x}^{\alpha} \dot{x}^{\beta}=\partial_{\beta} \Gamma_{\lambda \alpha}^{\mu} s^{\lambda} \dot{x}^{\alpha} \dot{x}^{\beta} and that Gamma_(alpha beta)^(mu)=Gamma_(beta alpha)^(mu)\Gamma_{\alpha \beta}^{\mu}=\Gamma_{\beta \alpha}^{\mu}. Thus, we can identify the Riemann curvature tensor as
b) In the Newtonian limit, we can approximate the metric by g_(mu nu)=eta_(mu nu)+g_{\mu \nu}=\eta_{\mu \nu}+h_(mu nu)h_{\mu \nu}, where |h_(mu nu)|≪1\left|h_{\mu \nu}\right| \ll 1 is small, and therefore, we have the metric tensor g_(mu nu)g_{\mu \nu}, the Christoffel symbols Gamma_(mu nu)^(lambda)\Gamma_{\mu \nu}^{\lambda}, and the Riemann curvature tensor R^(mu)_(nu lambda rho)R^{\mu}{ }_{\nu \lambda \rho} as
Furthermore, in the Newtonian limit, we have the static field condition, i.e., del_(0)g_(alpha beta)=0\partial_{0} g_{\alpha \beta}=0, which means that all time derivatives of the metric can be neglected, and thus, we find that
which, using the result found in a) with tau=t\tau=t, setting c=1c=1, and again choosing mu=i\mu=i, leads to the equation for the tidal acceleration in Newton's theory of gravitation, namely
which is what we wanted to show. In words, the acceleration of the separation between the trajectories of the two particles is given by a tensor of the second derivatives of the gravitational potential.
2.115
a) The metric in Minkowski space in Cartesian spatial coordinates is given by
{:(3.1810)ds^(2)=c^(2)dt^(2)-dx_(0)^(2)-dy_(0)^(2)-dz_(0)^(2):}\begin{equation*}
d s^{2}=c^{2} d t^{2}-d x_{0}^{2}-d y_{0}^{2}-d z_{0}^{2} \tag{3.1810}
\end{equation*}
Consider the motion of a free particle in the lab frame described by a cylindrical coordinate system x=r cos varphi,y=r sin varphix=r \cos \varphi, y=r \sin \varphi, and z=zz=z, but also in the rest frame of the particle that is rotating with constant angular velocity omega\omega around the zz-axis relative to the lab frame. The spatial cylindrical coordinates in the rest frame of the particle are given by x=r_(0)cos varphi_(0),y=r_(0)sin varphi_(0)x=r_{0} \cos \varphi_{0}, y=r_{0} \sin \varphi_{0}, and z=z_(0)z=z_{0} and the two frames are related to each other by the following coordinate transformations
Without loss of generality, we can set c=1c=1. In the rest frame of the particle, the metric becomes
{:[ds^(2)=dt^(2)-dr_(0)^(2)-r_(0)^(2)dvarphi_(0)^(2)-dz_(0)^(2)=dt^(2)-dr^(2)-r^(2)(d varphi-omega dt)^(2)-dz^(2)],[(3.1812)=(1-omega^(2))dt^(2)-dr^(2)-r^(2)dvarphi^(2)-dz^(2)+2r^(2)omega dtd varphi]:}\begin{align*}
d s^{2} & =d t^{2}-d r_{0}^{2}-r_{0}^{2} d \varphi_{0}^{2}-d z_{0}^{2}=d t^{2}-d r^{2}-r^{2}(d \varphi-\omega d t)^{2}-d z^{2} \\
& =\left(1-\omega^{2}\right) d t^{2}-d r^{2}-r^{2} d \varphi^{2}-d z^{2}+2 r^{2} \omega d t d \varphi \tag{3.1812}
\end{align*}
which in turn implies the Lagrangian
{:(3.1813)L=g_(mu nu)dx^(mu)dx^(nu)=(1-omega^(2)r^(2))t^(˙)^(2)+2omegar^(2)t^(˙)varphi^(˙)-r^(˙)^(2)-r^(2)varphi^(˙)^(2)-z^(˙)^(2):}\begin{equation*}
\mathcal{L}=g_{\mu \nu} d x^{\mu} d x^{\nu}=\left(1-\omega^{2} r^{2}\right) \dot{t}^{2}+2 \omega r^{2} \dot{t} \dot{\varphi}-\dot{r}^{2}-r^{2} \dot{\varphi}^{2}-\dot{z}^{2} \tag{3.1813}
\end{equation*}
Thus, we can immediately read off the nonzero components of the metric, namely
Now, using the Euler-Lagrange equations, i.e., (d)/(d tau)(delL)/(delx^(˙)^(mu))-(delL)/(delx^(mu))=0\frac{d}{d \tau} \frac{\partial \mathcal{L}}{\partial \dot{x}^{\mu}}-\frac{\partial \mathcal{L}}{\partial x^{\mu}}=0, we find that
Since (delL)/(del t)=(delL)/(del varphi)=(delL)/(del z)=0\frac{\partial \mathcal{L}}{\partial t}=\frac{\partial \mathcal{L}}{\partial \varphi}=\frac{\partial \mathcal{L}}{\partial z}=0, we basically have three constants of motion, i.e., (delL)/(del t)=\frac{\partial \mathcal{L}}{\partial t}= const., (delL)/(del(varphi^(˙)))=\frac{\partial \mathcal{L}}{\partial \dot{\varphi}}= const., and (delL)/(del z)=\frac{\partial \mathcal{L}}{\partial z}= const. However, we observe that two of the EulerLagrange equations contain both t^(¨)\ddot{t} and varphi^(¨)\ddot{\varphi}, and therefore, solving for t^(¨)\ddot{t} and varphi^(¨)\ddot{\varphi}, we obtain the geodesic equations
Thus, comparing the geodesic equations with the general formula for the geodesic equations, i.e., x^(¨)^(mu)+Gamma_(nu lambda)^(mu)x^(˙)^(nu)x^(˙)^(lambda)=0\ddot{x}^{\mu}+\Gamma_{\nu \lambda}^{\mu} \dot{x}^{\nu} \dot{x}^{\lambda}=0, and using Gamma_(nu lambda)^(mu)=Gamma_(lambda nu)^(mu)\Gamma_{\nu \lambda}^{\mu}=\Gamma_{\lambda \nu}^{\mu}, we can read off the eight nonzero Christoffel symbols as
Note that in the case we set omega=0\omega=0, the Christoffel symbols reduce to three, i.e., Gamma_(varphi varphi)^(r)=-r\Gamma_{\varphi \varphi}^{r}=-r and Gamma_(r varphi)^(varphi)=Gamma_(varphi r)^(varphi)=(1)/(r)\Gamma_{r \varphi}^{\varphi}=\Gamma_{\varphi r}^{\varphi}=\frac{1}{r}, which are the Christoffel symbols in nonrotating cylinder coordinates.
To summarize, in the rest frame of a free particle rotating with constant angular velocity omega\omega around the zz-coordinate axis, the Christoffel symbols and the geodesic equations are
since for small velocities, the component x^(˙)^(0)(sigma)\dot{x}^{0}(\sigma) of the 4 -velocity is much larger than the corresponding spatial components, and where sigma\sigma is the curve parameter. Using the results found in a), we only have one nonzero Christoffel symbol on the form Gamma_(00)^(mu)\Gamma_{00}^{\mu}, which is Gamma_(00)^(r)=Gamma_(tt)^(r)=-omega^(2)r\Gamma_{00}^{r}=\Gamma_{t t}^{r}=-\omega^{2} r, and therefore, the approximate geodesic equations are reduced to
In the rest frame of the particle, the first equation means that we can choose the time tt as the curve parameter sigma\sigma, i.e., t=sigmat=\sigma, and then the second equation becomes
Multiplying this equation with the mass mm of the particle, we obtain
{:(3.1825)mr^(¨)=momega^(2)r-=F_("centrifugal ")",":}\begin{equation*}
m \ddot{r}=m \omega^{2} r \equiv F_{\text {centrifugal }}, \tag{3.1825}
\end{equation*}
where F_("centrifugal ")=momega^(2)rF_{\text {centrifugal }}=m \omega^{2} r is the centrifugal force that appears to act on all objects when viewed in a rotating frame.
Thus, in the nonrelativistic limit, Newton's equation of motion in the rotating frame of the free particle is given by
{:(3.1826)mr^(¨)=momega^(2)r.:}\begin{equation*}
m \ddot{r}=m \omega^{2} r . \tag{3.1826}
\end{equation*}
Note that this equation is expressed in the radial coordinate of either the lab frame or the rest frame of the particle, since r=r_(0)r=r_{0} [see a)].
2.116
a) In spherical coordinates the Lagrangian is
{:(3.1827)L=(1)/(2)mr^(˙)^(2)+(GMm)/(r)=(1)/(2)m(r^(˙)^(2)+r^(2)varphi^(˙)^(2))+(GMm)/(r):}\begin{equation*}
\mathcal{L}=\frac{1}{2} m \dot{\mathbf{r}}^{2}+\frac{G M m}{r}=\frac{1}{2} m\left(\dot{r}^{2}+r^{2} \dot{\varphi}^{2}\right)+\frac{G M m}{r} \tag{3.1827}
\end{equation*}
where it is assumed that theta=pi//2\theta=\pi / 2 and theta^(˙)=0\dot{\theta}=0 with dot meaning differentiation with respect to time tt. This yields the Euler-Lagrange equations
where it holds that r^(˙)=r^(¨)=0\dot{r}=\ddot{r}=0, since it is assumed that r=r_(0)=r=r_{0}= const. The second equation is solved by L=mr^(2)varphi^(˙)=L=m r^{2} \dot{\varphi}= constant, i.e., varphi^(˙)=L//(mr^(2))\dot{\varphi}=L /\left(m r^{2}\right). Inserting this into the first equation, we obtain
{:(3.1830)-mr((L)/(mr^(2)))^(2)+(GMm)/(r^(2))=0=>r=(L^(2))/(GMm^(2))-=r_(0):}\begin{equation*}
-m r\left(\frac{L}{m r^{2}}\right)^{2}+\frac{G M m}{r^{2}}=0 \Rightarrow r=\frac{L^{2}}{G M m^{2}} \equiv r_{0} \tag{3.1830}
\end{equation*}
which is the trajectory of the planet.
Note that r_(0)r_{0} is exactly the radius where the centrifugal force mr^(2)varphi^(˙)m r^{2} \dot{\varphi} is balanced by the attractive gravitational force GMm//r^(2)G M m / r^{2}, i.e., mr^(2)varphi^(˙)=GMm//r^(2)m r^{2} \dot{\varphi}=G M m / r^{2}, which together with L=mr^(2)varphi^(˙)L=m r^{2} \dot{\varphi} gives an elementary derivation of the trajectory of the planet.
b) The result in a) can be summarized as follows: It has been shown that mr^(¨)=-mPhi_("eff ")^(')(r)m \ddot{r}=-m \Phi_{\text {eff }}^{\prime}(r) with the effective gravitational potential
{:(3.1831)mPhi_(eff)(r)=-(GMm)/(r)+(1)/(2)mr^(2)varphi^(˙)^(2)=m(-(GM)/(r)+(L^(2))/(2m^(2)r^(2))):}\begin{equation*}
m \Phi_{\mathrm{eff}}(r)=-\frac{G M m}{r}+\frac{1}{2} m r^{2} \dot{\varphi}^{2}=m\left(-\frac{G M}{r}+\frac{L^{2}}{2 m^{2} r^{2}}\right) \tag{3.1831}
\end{equation*}
The stationary orbit r=r_(0)r=r_{0} can thus be obtained as the solution to Phi_("eff ")^(')(r)=0\Phi_{\text {eff }}^{\prime}(r)=0. This solution r=r_(0)r=r_{0} is a minimum of Phi_("eff ")(r)\Phi_{\text {eff }}(r), and therefore, it corresponds to a stable orbit.
There exists a natural generalization of the effective potential Phi_("eff ")(r)\Phi_{\text {eff }}(r) to general relativity, namely
where r_(0)-=L^(2)//(GMm^(2))r_{0} \equiv L^{2} /\left(G M m^{2}\right), which is the solution to the problem in a). The general relativity generalization of the orbit r_(0)r_{0} can be computed by solving Phi^(')(r)=0\Phi^{\prime}(r)=0, which yields
The solution r_(+)r_{+}corresponds to a minimum of Phi_("eff ")(r)\Phi_{\text {eff }}(r) and is thus a stable orbit generalizing the Newtonian solution found in a), whereas the solution r_(-)r_{-}corresponds to a maximum and is thus unstable, but r_(-)rarr0r_{-} \rightarrow 0 in the Newtonian limit r_(**)rarr0r_{*} \rightarrow 0. Therefore, the generalization of the orbit r_(0)r_{0} in a) to general relativity is
where eta_(mu nu)\eta_{\mu \nu} is the Minkowski metric and h_(mu nu)h_{\mu \nu} is a small perturbation. We will only need to determine the Riemann tensor to first order in h_(mu nu)h_{\mu \nu}. Furthermore, we assume that the source of the gravitational field is not large and slowly moving, hence only the component T^(00)=c^(2)rhoT^{00}=c^{2} \rho gives a relevant contribution to the field. Furthermore, we assume that in the Newtonian limit rho≫(p)/(c^(2))\rho \gg \frac{p}{c^{2}} hence we can assume T^(mu nu)=rhoU^(mu)U^(nu)T^{\mu \nu}=\rho U^{\mu} U^{\nu}.
b) Einstein's equations are
where G^(mu nu)=R^(mu nu)-(1)/(2)Rg^(mu nu)G^{\mu \nu}=R^{\mu \nu}-\frac{1}{2} R g^{\mu \nu}. For the metric in g_(mu nu)=eta_(mu nu)+h_(mu nu)g_{\mu \nu}=\eta_{\mu \nu}+h_{\mu \nu} the Christoffel symbols are given by
Since the lowest order of h_(mu nu)h_{\mu \nu} in the Christoffel symbols is the first, the Riemann tensor of first order will therefore only contain the derivative terms
Using the gauge condition del^(lambda)h_(mu lambda)=del_(mu)h//2\partial^{\lambda} h_{\mu \lambda}=\partial_{\mu} h / 2, we can write the Ricci tensor and Ricci scalar as
we find grad^(2)phi=4pi G rho\nabla^{2} \phi=4 \pi G \rho.
2.118
The metric outside of the Earth is given by
{:(3.1846)ds^(2)=(1+2Phi)dt^(2)+(1+2Phi)^(-1)dr^(2)-r^(2)dOmega^(2):}\begin{equation*}
d s^{2}=(1+2 \Phi) d t^{2}+(1+2 \Phi)^{-1} d r^{2}-r^{2} d \Omega^{2} \tag{3.1846}
\end{equation*}
where Phi=-(GM)/(r)\Phi=-\frac{G M}{r} is the gravitational potential and dOmega^(2)=dtheta^(2)+sin^(2)theta dphi^(2)d \Omega^{2}=d \theta^{2}+\sin ^{2} \theta d \phi^{2}. A satellite is orbiting at a distance R_(1)R_{1} from the surface. We are interested the eigentime for the satellite to complete a full orbit around the Earth. Therefore, we have R=R_(E)+R_(1)R=R_{E}+R_{1} and due to the spherical symmetry we can assume theta=(pi)/(2)\theta=\frac{\pi}{2}. Thus, the metric can be written as
{:(3.1847)ds^(2)=(1+2Phi)dt^(2)-(1+2Phi)^(-1)dr^(2)-R^(2)dphi^(2):}\begin{equation*}
d s^{2}=(1+2 \Phi) d t^{2}-(1+2 \Phi)^{-1} d r^{2}-R^{2} d \phi^{2} \tag{3.1847}
\end{equation*}
where alpha\alpha and beta\beta are constants. From the first equation, we can determine a relation between t^(˙)\dot{t} and phi^(˙)\dot{\phi}, using r^(˙)=0\dot{r}=0 and r^(¨)=0\ddot{r}=0. Inserting the expressions for t^(˙)\dot{t} and phi^(˙)\dot{\phi}, we get
{:(3.1854)Rbeta^(2)-(GM)/(R^(2))alpha^(2)=0:}\begin{equation*}
R \beta^{2}-\frac{G M}{R^{2}} \alpha^{2}=0 \tag{3.1854}
\end{equation*}
Furthermore, the Lagrangian can be reduced, using r^(˙)=0\dot{r}=0, to
which can be solved for alpha=(1)/(sqrt(1+(3)/(2)*Phi))\alpha=\frac{1}{\sqrt{1+\frac{3}{2} \cdot \Phi}} and beta=sqrt((Phi)/(R(2+3Phi)))\beta=\sqrt{\frac{\Phi}{R(2+3 \Phi)}}, and thus, from (d phi)/(d tau)=beta\frac{d \phi}{d \tau}=\beta, we get
Figure 3.17 Setup of the spherical body of radius R_(0)R_{0} and impact parameter bb and deflection angle alpha\alpha of the neutrino, respectively.
2.119
See Figure 3.17 for the setup of the problem. The deflection angle due to gravitational lensing in the weak-field limit is given by
{:(3.1858)alpha=|2int(grad phi)_(_|_)dt|:}\begin{equation*}
\alpha=\left|2 \int(\nabla \phi)_{\perp} d t\right| \tag{3.1858}
\end{equation*}
where phi\phi is the Newtonian gravitational potential and (grad phi)_(_|_)(\nabla \phi)_{\perp} is the component of its gradient which is perpendicular to the zeroth order direction of motion. We pick a coordinate system such that the zeroth order worldline of the neutrinos is given by x(t)=te_(1)+be_(2)x(t)=t e_{1}+b e_{2}. It follows that
{:(3.1859)alpha=2|int_(-oo)^(oo)e_(2)*grad phi dt|=4|int_(0)^(oo)e_(2)*grad phi dt|:}\begin{equation*}
\alpha=2\left|\int_{-\infty}^{\infty} e_{2} \cdot \nabla \phi d t\right|=4\left|\int_{0}^{\infty} e_{2} \cdot \nabla \phi d t\right| \tag{3.1859}
\end{equation*}
where M(r)M(r) is the mass inside the radius rr. This mass is given by
{:(3.1861)M(r)={[M_(0)(r^(3))/(R_(0)^(3))","quad r < r_(0)],[M_(0)","quad r >= r_(0)]:}:}M(r)=\left\{\begin{array}{cc}
M_{0} \frac{r^{3}}{R_{0}^{3}}, \quad r<r_{0} \tag{3.1861}\\
M_{0}, \quad r \geq r_{0}
\end{array}\right.
Noting that r^(2)=t^(2)+b^(2)r^{2}=t^{2}+b^{2}, it follows that
{:[int_(0)^(oo)e_(2)*grad phi dt=int_(0)^(oo)(GbM(r))/(r^(3))dt=GM_(0)int_(0)^(sqrt(R_(0)^(2)-b^(2)))(b)/(R_(0)^(3))dt+GM_(0)int_(sqrt(R_(0)^(2)-b^(2)))^(oo)(b)/(r^(3))dt],[=M_(0)bG[(sqrt(R_(0)^(2)-b^(2)))/(R_(0)^(3))+(1)/(b^(2))-(1)/(b^(2))*(sqrt(R_(0)^(2)-b^(2)))/(R_(0))]],[(3.1862)=M_(0)bG[(sqrt(R_(0)^(2)-b^(2)))/(R_(0))((1)/(R_(0)^(2))-(1)/(b^(2)))+(1)/(b^(2))]]:}\begin{align*}
\int_{0}^{\infty} e_{2} \cdot \nabla \phi d t & =\int_{0}^{\infty} \frac{G b M(r)}{r^{3}} d t=G M_{0} \int_{0}^{\sqrt{R_{0}^{2}-b^{2}}} \frac{b}{R_{0}^{3}} d t+G M_{0} \int_{\sqrt{R_{0}^{2}-b^{2}}}^{\infty} \frac{b}{r^{3}} d t \\
& =M_{0} b G\left[\frac{\sqrt{R_{0}^{2}-b^{2}}}{R_{0}^{3}}+\frac{1}{b^{2}}-\frac{1}{b^{2}} \cdot \frac{\sqrt{R_{0}^{2}-b^{2}}}{R_{0}}\right] \\
& =M_{0} b G\left[\frac{\sqrt{R_{0}^{2}-b^{2}}}{R_{0}}\left(\frac{1}{R_{0}^{2}}-\frac{1}{b^{2}}\right)+\frac{1}{b^{2}}\right] \tag{3.1862}
\end{align*}
This leads us to conclude that the deflection angle alpha\alpha is given by
{:(3.1863)alpha=4M_(0)bG[(sqrt(R_(0)^(2)-b^(2)))/(R_(0))((1)/(R_(0)^(2))-(1)/(b^(2)))+(1)/(b^(2))]:}\begin{equation*}
\alpha=4 M_{0} b G\left[\frac{\sqrt{R_{0}^{2}-b^{2}}}{R_{0}}\left(\frac{1}{R_{0}^{2}}-\frac{1}{b^{2}}\right)+\frac{1}{b^{2}}\right] \tag{3.1863}
\end{equation*}
For b rarrR_(0),alpha rarr(4M_(0)G)/(b)b \rightarrow R_{0}, \alpha \rightarrow \frac{4 M_{0} G}{b}, which is the expected result as it coincides with the result for the deflection outside of a spherical mass distribution with total mass M_(0)M_{0}.
2.120
The Newtonian gravitational potential Phi\Phi satisfies Poisson's equation
{:(3.1864)grad^(2)Phi=4pi G rho:}\begin{equation*}
\nabla^{2} \Phi=4 \pi G \rho \tag{3.1864}
\end{equation*}
in the region r > r_(s)r>r_{s}. Due to the symmetry of the problem, we will assume that Phi\Phi is spherically symmetric and introduce the function f(r)=r Phi(r)f(r)=r \Phi(r), with which the differential equation for the gravitational potential takes the form
{:(3.1865)f^('')(r)=4pi rG rho(r):}\begin{equation*}
f^{\prime \prime}(r)=4 \pi r G \rho(r) \tag{3.1865}
\end{equation*}
With the assumption of the NFW halo profile for rho(r)\rho(r), we find that
{:(3.1866)f^('')(r)=(4pi Gk)/(r^(2))quad Longrightarrowquad f(r)=-4pi Gk ln((r)/(R))+Cr:}\begin{equation*}
f^{\prime \prime}(r)=\frac{4 \pi G k}{r^{2}} \quad \Longrightarrow \quad f(r)=-4 \pi G k \ln \left(\frac{r}{R}\right)+C r \tag{3.1866}
\end{equation*}
where CC and RR are integration constants. The resulting expression for the gravitational potential is therefore
{:(3.1867)Phi(r)=-(4pi Gk)/(r)ln((r)/(R))+C:}\begin{equation*}
\Phi(r)=-\frac{4 \pi G k}{r} \ln \left(\frac{r}{R}\right)+C \tag{3.1867}
\end{equation*}
The gradient of the potential is now
{:(3.1868)grad Phi=e_(r)Phi^(')(r)=(4pi Gk)/(r^(2))ln((r)/(eR))=e_(r)[(4pi Gk)/(r^(2))ln((r)/(r_(s)))+(K)/(r^(2))]:}\begin{equation*}
\nabla \Phi=\mathbf{e}_{r} \Phi^{\prime}(r)=\frac{4 \pi G k}{r^{2}} \ln \left(\frac{r}{e R}\right)=\mathbf{e}_{r}\left[\frac{4 \pi G k}{r^{2}} \ln \left(\frac{r}{r_{s}}\right)+\frac{K}{r^{2}}\right] \tag{3.1868}
\end{equation*}
where we have introduced the new constant K=4pi Gk ln(r_(s)//eR)K=4 \pi G k \ln \left(r_{s} / e R\right). The constant KK can be related to the mass M_(0)M_{0} enclosed within r < r_(s)r<r_{s} according to the relation
Letting the zeroth order worldline of the light signal be given by x(t)=te_(1)+r_(0)e_(2)\mathbf{x}(t)=t \mathbf{e}_{1}+r_{0} \mathbf{e}_{2}, the first-order lensing is given by
{:(3.1870)theta≃2inte_(2)*grad Phi dt=2int_(-oo)^(oo)(GM_(0)r_(0))/(sqrt(r_(0)^(2)+t^(2)))dt+4pi Gkint_(-oo)^(oo)(ln[(t^(2)+r_(0)^(2))//r_(s)^(2)])/(sqrt(r_(0)^(2)+t^(2))^(3))dt:}\begin{equation*}
\theta \simeq 2 \int \mathbf{e}_{2} \cdot \nabla \Phi d t=2 \int_{-\infty}^{\infty} \frac{G M_{0} r_{0}}{\sqrt{r_{0}^{2}+t^{2}}} d t+4 \pi G k \int_{-\infty}^{\infty} \frac{\ln \left[\left(t^{2}+r_{0}^{2}\right) / r_{s}^{2}\right]}{{\sqrt{r_{0}^{2}+t^{2}}}^{3}} d t \tag{3.1870}
\end{equation*}
The first of these integrals is the same integral that we have encountered for the gravitational lensing of a point source. The second integral can be solved through a lengthy process involving substitutions and partial integrations to give the final result
{:(3.1871)theta≃(4GM_(0))/(r_(0))+(16 pi Gk)/(r_(0))ln((er_(0))/(2r_(s)))≃(4GM_(0))/(r_(0))+(16 pi Gk)/(r_(0))[ln((r_(0))/(r_(s)))+0.31]:}\begin{equation*}
\theta \simeq \frac{4 G M_{0}}{r_{0}}+\frac{16 \pi G k}{r_{0}} \ln \left(\frac{e r_{0}}{2 r_{s}}\right) \simeq \frac{4 G M_{0}}{r_{0}}+\frac{16 \pi G k}{r_{0}}\left[\ln \left(\frac{r_{0}}{r_{s}}\right)+0.31\right] \tag{3.1871}
\end{equation*}
where e=2.71828 dotse=2.71828 \ldots is the base of the natural logarithm.
2.121
a) Consider two observers AA and BB located at spacetime points x_(A)^(mu)x_{A}^{\mu} and x_(B)^(mu)x_{B}^{\mu}, respectively. Suppose that AA is sending out light waves at the rate nn per coordinate time interval Delta\Delta. In the (local) rest frame of an observer, the time interval Delta\Delta is related to the proper time interval by the factor sqrt(g_(00))\sqrt{g_{00}}. It follows that the relation between the number of light waves received by BB per unit proper time and the number of light waves emitted by AA per unit proper time is
where v_(A)(lambda_(A))v_{A}\left(\lambda_{A}\right) is the frequency (wavelength) of the light emitted by AA and v_(B)(lambda_(B))v_{B}\left(\lambda_{B}\right) is the frequency (wavelength) of the light received by BB. Note that a frequency vv and its corresponding wavelength lambda\lambda are related as v lambda=cv \lambda=c. The light is traveling along a null geodesic (i.e., a geodesic such that the tangent vector at each point is lightlike) from AA to BB, but we do not need the explicit solution of the geodesic equations.
b) In general, if the radial coordinate of BB is r_(B)≫r_(**)r_{B} \gg r_{*} in the Schwarzschild spacetime with g_(00)=1-r_(**)//rg_{00}=1-r_{*} / r, then
This means that the frequency observed by BB is actually smaller than the frequency emitted by AA and the wavelength observed by BB is longer than the wavelength emitted by AA, i.e., the light is redshifted. The redshift zz is defined as z-=lambda_(B)//lambda_(A)-1z \equiv \lambda_{B} / \lambda_{A}-1.
Finally, assume that light is emitted by AA at r_(A)r_{A} (e.g., from the surface of a star) and the light is observed by BB far away (on Earth), i.e., r_(B)-=r_(oo)≫r_(**)r_{B} \equiv r_{\infty} \gg r_{*}, then the formula for the gravitational redshift is given by
Note that for light emitted at r rarrr_(**)r \rightarrow r_{*}, the redshift grows to infinity, whereas for light emitted from r_(A)≫r_(**)r_{A} \gg r_{*}, there is no redshift. In the Newtonian limit, when r_(A)r_{A} is sufficiently large compared to r_(**)r_{*}, i.e., r_(A)≫r_(**)r_{A} \gg r_{*}, the gravitational redshift can be approximated as
White dwarfs like Sirius B and 40 Eridani B do show gravitational redshifts in the range between 10^(-4)10^{-4} and 10^(-5)10^{-5}, which are of the right order of magnitude. More reliable and quantitatively accurate measurements are possible only in terrestrial experiments. For example, in 1960, Pound and Rebka measured the change of frequency of a gamma\gamma-ray photon emitted by an excited iron nucleus as it fell from a height of 18-21m18-21 \mathrm{~m}. When the photon falls from a height hh, the change in the Newtonian gravitational potential is ghg h, where gg is the acceleration due to gravity on the Earth's surface. Since g_(00)-1g_{00}-1 is approximately given by the Newtonian gravitational potential, the frequency increases by a factor 1+gh//c^(2)1+g h / c^{2}. The fraction gh//c^(2)g h / c^{2} is small, about 10^(-15)10^{-15}, but it can still be measured, confirming the gravitational redshift effect.
2.122
The surface (ct)^(2)-x^(2)-y^(2)=-K^(2)(c t)^{2}-x^{2}-y^{2}=-K^{2}, where K > 0K>0, can be written as (ct)^(2)-r^(2)=(c t)^{2}-r^{2}=-K^(2)-K^{2} if we introduce polar coordinates such that (t,x,y)=(t,r cos phi,r sin phi)(t, x, y)=(t, r \cos \phi, r \sin \phi). Thus, we have
Therefore, the position vector is (t,x,y)=(t,sqrt(c^(2)t^(2)+K^(2))cos phi,sqrt(c^(2)t^(2)+K^(2))sin phi)(t, x, y)=\left(t, \sqrt{c^{2} t^{2}+K^{2}} \cos \phi, \sqrt{c^{2} t^{2}+K^{2}} \sin \phi\right). Differentiating the position vector, we obtain
Since the metric components are independent of phi\phi, the vector field del_(phi)\partial_{\phi} is a Killing vector field and with N=alphadel_(t)+betadel_(phi)N=\alpha \partial_{t}+\beta \partial_{\phi} being the 4 -frequency of a light pulse sent from AA at time coordinate tt to BB at time coordinate t^(')t^{\prime}, this implies that
For any comoving observer with constant phi\phi, the 4 -velocity of the observer is given by V=gammadel_(t)V=\gamma \partial_{t}. Normalizing this to one we obtain
where we have used that g_(rr)=-1//g_(00)g_{r r}=-1 / g_{00} and that the light is moving radially toward smaller rr. When the light signal is emitted from r rarr oor \rightarrow \infty, we have g_(00)rarr1g_{00} \rightarrow 1 and thus E=omega_(0)E=\omega_{0}, where omega_(0)\omega_{0} is the frequency observed by a stationary observer at infinity.
The 4-frequency is therefore generally given by
Using that lambda prop1//omega\lambda \propto 1 / \omega and inserting g_(00)=1-2GM//rg_{00}=1-2 G M / r the yields
{:(3.1895)lambda=lambda_(0)sqrt(1-(2GM)/(r))sqrt((1-(2GM)/(r)-v_(c))/(1-(2GM)/(r)+v_(c))):}\begin{equation*}
\lambda=\lambda_{0} \sqrt{1-\frac{2 G M}{r}} \sqrt{\frac{1-\frac{2 G M}{r}-v_{c}}{1-\frac{2 G M}{r}+v_{c}}} \tag{3.1895}
\end{equation*}
Inserting the values provided in the problem, we find that
The gravitational redshift observed far away for light emitted from the surface of a star (when the mass of the star is M=2*10^(30)kgM=2 \cdot 10^{30} \mathrm{~kg} and r_("star ")=7*10^(8)mr_{\text {star }}=7 \cdot 10^{8} \mathrm{~m} ) is given by
See Problem 2.121 for a derivation of the formula for the gravitational redshift.
2.125
The Schwarzschild metric is given by
{:(3.1898)ds^(2)=(1-(2GM)/(c^(2)r))(dx^(0))^(2)-(1-(2GM)/(c^(2)r))^(-1)dr^(2)-r^(2)dOmega^(2):}\begin{equation*}
d s^{2}=\left(1-\frac{2 G M}{c^{2} r}\right)\left(d x^{0}\right)^{2}-\left(1-\frac{2 G M}{c^{2} r}\right)^{-1} d r^{2}-r^{2} d \Omega^{2} \tag{3.1898}
\end{equation*}
From this metric, one obtains
{:(3.1899)g_(00)(r)=1-(2GM)/(c^(2)r):}\begin{equation*}
g_{00}(r)=1-\frac{2 G M}{c^{2} r} \tag{3.1899}
\end{equation*}
Now, one finds the ratio between the observed frequency v^(')v^{\prime} and the emitted frequency vv as (see the discussion in the solution to Problem 2.121)
{:(3.1900)(v^('))/(v)=sqrt((g_(00)(r))/(g_(00)(r^('))))=sqrt((1-(2GM)/(c^(2)r))/(1-(2GM)/(c^(2)r^(')))):}\begin{equation*}
\frac{v^{\prime}}{v}=\sqrt{\frac{g_{00}(r)}{g_{00}\left(r^{\prime}\right)}}=\sqrt{\frac{1-\frac{2 G M}{c^{2} r}}{1-\frac{2 G M}{c^{2} r^{\prime}}}} \tag{3.1900}
\end{equation*}
where rr is the solar radius and r^(')r^{\prime} is the average Sun-Earth distance.
Then, since r^(')≫rr^{\prime} \gg r and r≫2GM//c^(2)r \gg 2 G M / c^{2}, one has
When the spaceship is sending the light signal to the Earth, there are two effects: One is the redshift due to the Doppler effect, the other is a blueshift due to the gravitational pull from the Earth.
Let the frequency of the emitted light be nu\nu. The redshift is then given by the zz-factor, which is
where v_("obs ")v_{\text {obs }} is the observed frequency and vv is the velocity of the spaceship. Using v=100m//sv=100 \mathrm{~m} / \mathrm{s} and c=3*10^(8)m//sc=3 \cdot 10^{8} \mathrm{~m} / \mathrm{s}, we obtain for the redshift z_(r)~~3.3*10^(-7)z_{r} \approx 3.3 \cdot 10^{-7}.
For the blueshift due the the mass of the Earth, we use the Schwarzschild metric.
Thus, we have
{:(3.1905)z_(b)=(v)/(v_(obs))-1=sqrt((g_(00)(R))/(g_(00)(R+h)))-1","quadg_(00)(r)=1-(2GM)/(c^(2)r):}\begin{equation*}
z_{b}=\frac{v}{v_{\mathrm{obs}}}-1=\sqrt{\frac{g_{00}(R)}{g_{00}(R+h)}}-1, \quad g_{00}(r)=1-\frac{2 G M}{c^{2} r} \tag{3.1905}
\end{equation*}
where RR is the radius of the Earth and hh is the altitude of the spaceship. Since h//Rh / R(~~0.16≪1)(\approx 0.16 \ll 1) is small, we obtain
where gg is the acceleration at the Earth, i.e., g≃9.8m//s^(2)g \simeq 9.8 \mathrm{~m} / \mathrm{s}^{2}. Since h=10^(6)mh=10^{6} \mathrm{~m}, we obtain for the blueshift z_(b)~~-1.1*10^(-10)z_{b} \approx-1.1 \cdot 10^{-10}.
In conclusion, the redshift is much larger (i.e., 3000 times) than the blueshift. Thus, in this case, the Doppler effect is the most important physical effect.
2.127
The gravitational potential of the Sun at the Earth is given by
which is an order of magnitude larger than the gravitational potential of the Earth at its surface. Thus, we can neglect the influence of the gravitational field of the Earth. The relation between the frequencies at (v_(o+))\left(v_{\oplus}\right) and far away from (v_(oo))\left(v_{\infty}\right) the Earth will be
for small phi_(o.)//c^(2)\phi_{\odot} / c^{2}. See the general discussion of this relation in the solution to Problem 2.125. The redshift parameter zz is given by
We call the free-falling observer AA and the observer at infinity BB. The motion of the free-falling observer is governed by the radial differential equation
where EE is a constant of motion given by sqrt(2E)=g(del_(t),U)\sqrt{2 E}=g\left(\partial_{t}, U\right), where UU is the 4 -velocity of the observer AA. Note that the constant of motion g(del_(varphi),U)=0g\left(\partial_{\varphi}, U\right)=0, since the motion of AA is assumed to be purely radial. In order to have a velocity just large enough to escape to infinity, we require that r^(˙)rarr0\dot{r} \rightarrow 0 as r rarr oor \rightarrow \infty, leading to E=1//2E=1 / 2 and therefore
Furthermore, we know that NN is tangent to and parallel along the lightlike geodesic describing the worldline of the light signal. Since del_(t)\partial_{t} is a Killing vector field of the Schwarzschild spacetime, we find that
Light signals follow geodesics and their 4-frequency NN is proportional to the tangent of the affinely parametrized geodesic. Since K=del_(t)K=\partial_{t} is a Killing vector field, we know that
is constant along its worldline gamma_(0)\gamma_{0}. Furthermore, g(gamma^(˙)_(0),gamma^(˙)_(0))=(Q^(2))/(x^(2))-x^(˙)^(2)=1g\left(\dot{\gamma}_{0}, \dot{\gamma}_{0}\right)=\frac{Q^{2}}{x^{2}}-\dot{x}^{2}=1. Since the observer falls from x=x_(0)x=x_{0}, we find that Q=+x_(0)Q=+x_{0}. We can conclude that t^(˙)=(x_(0))/(x^(2))\dot{t}=\frac{x_{0}}{x^{2}}, x^(˙)=-sqrt((x_(0)r)/(x^(2))-1)\dot{x}=-\sqrt{\frac{x_{0} r}{x^{2}}-1} for the falling observer. The emitted frequency f_(e)f_{e} is given by
The observer at x=x_(1)x=x_{1} with worldline gamma_(1)\gamma_{1} has x^(˙)=0\dot{x}=0 and therefore g(gamma^(˙)_(1),gamma^(˙)_(1))=g\left(\dot{\gamma}_{1}, \dot{\gamma}_{1}\right)=x_(1)^(2)t^(˙)^(2)=1x_{1}^{2} \dot{t}^{2}=1, which means that t^(˙)=x_(1)^(-1)\dot{t}=x_{1}^{-1}. The observed frequency is therefore
Consider the Robertson-Walker metric for d Omega=0d \Omega=0, i.e.,
{:(3.1928)ds^(2)=c^(2)dt^(2)-(S(t)^(2))/(1-kr^(2))dr^(2):}\begin{equation*}
d s^{2}=c^{2} d t^{2}-\frac{S(t)^{2}}{1-k r^{2}} d r^{2} \tag{3.1928}
\end{equation*}
Using ds^(2)=0d s^{2}=0, which holds for the path of a light signal, we find that
{:(3.1929)0=c^(2)dt^(2)-(S(t)^(2))/(1-kr^(2))dr^(2)quad=>quad(c)/(S(t))dt=(1)/(sqrt(1-kr^(2)))dr:}\begin{equation*}
0=c^{2} d t^{2}-\frac{S(t)^{2}}{1-k r^{2}} d r^{2} \quad \Rightarrow \quad \frac{c}{S(t)} d t=\frac{1}{\sqrt{1-k r^{2}}} d r \tag{3.1929}
\end{equation*}
where we assumed that dr//dt > 0d r / d t>0, i.e., propagation forward in time. Since the observers are at rest with respect to rr, we must have
where the dot indicates differentiation with respect to the path parameter ss. Using Euler-Lagrange equations yields the differential equations for the geodesics, i.e.,
Note that, since the term including the perturbations is linear in epsi\varepsilon already, the indices of xi^(a)\xi^{a} are can be raised and lowered with the Minkowski metric without affecting this statement and the lowering of the indices inside the partial derivatives is consistent to linear order in epsi\varepsilon. Thus, we are still in the weak-field regime after the coordinate change with the new metric perturbation given above. Furthermore, we find that
as long as both chi^(c)\chi^{c} and xi^(c)\xi^{c} are harmonic functions. The coordinate change therefore also preserves the harmonic gauge condition.
b) In the harmonic gauge, we find that
where we note that grad_(b)chi^(c)=del_(b)chi^(c)=delta_(b)^(c)\nabla_{b} \chi^{c}=\partial_{b} \chi^{c}=\delta_{b}^{c} and cc is just a counter for the coordinate functions, not a tensor index. In the weak-field limit, we also have
With xi^(a)=iC^(a)exp(ik*chi)\xi^{a}=i C^{a} \exp (i k \cdot \chi), where k*chi=k_(c)chi^(c)k \cdot \chi=k_{c} \chi^{c}, the partial derivatives of xi_(a)\xi_{a} are given by
{:(3.1947)del_(b)xi_(a)=eta_(ac)del_(b)xi^(c)=eta_(ac)iC^(c)ik_(b)exp(ik*chi)=-C_(a)k_(b)exp(ik*chi):}\begin{equation*}
\partial_{b} \xi_{a}=\eta_{a c} \partial_{b} \xi^{c}=\eta_{a c} i C^{c} i k_{b} \exp (i k \cdot \chi)=-C_{a} k_{b} \exp (i k \cdot \chi) \tag{3.1947}
\end{equation*}
d) From the harmonic gauge condition, we find that
{:(3.1949)del^(a) bar(h)_(ab)=A_(ab)ik^(a)exp(ik*chi)=0:}\begin{equation*}
\partial^{a} \bar{h}_{a b}=A_{a b} i k^{a} \exp (i k \cdot \chi)=0 \tag{3.1949}
\end{equation*}
i.e., with the given k_(0)=k_(3)=1k_{0}=k_{3}=1 (or, equivalently, k^(0)=-k^(3)=1k^{0}=-k^{3}=1 ),
for all bb. Due to the symmetry of AA, this also means that A_(b0)=A_(b3)A_{b 0}=A_{b 3}. This leads to A_(00)^(')=A_(03)^(')=A_(30)^(')=A_(33)^(')A_{00}^{\prime}=A_{03}^{\prime}=A_{30}^{\prime}=A_{33}^{\prime} and it is therefore sufficient to require that A_(00)^(')=0A_{00}^{\prime}=0. Furthermore, we also find that
a) Consider the quantities h_(mu nu)h_{\mu \nu} to be perturbations or deviations of the components of the metric tensor away from flat spacetime. In general, for h_(mu nu)h_{\mu \nu}, the Lorenz gauge condition is given by
Cf. the analogy to this gauge condition in electromagnetism. If the Lorenz gauge condition is fulfilled, then Einstein's equations in vacuum reduce to ◻ bar(h)_(mu nu)=0\square \bar{h}_{\mu \nu}=0, where bar(h)_(mu nu)=h_(mu nu)-eta_(mu nu)h//2\bar{h}_{\mu \nu}=h_{\mu \nu}-\eta_{\mu \nu} h / 2 with h=eta^(mu nu)h_(mu nu)h=\eta^{\mu \nu} h_{\mu \nu}. This gauge condition does not completely fix the coordinates, so in order to do so, further conditions can be imposed, e.g., h_(i0)=0h_{i 0}=0 and eta^(mu nu)h_(mu nu)=0\eta^{\mu \nu} h_{\mu \nu}=0, which are collectively referred to as the transverse traceless gauge (or TT gauge), where h_(mu nu)= bar(h)_(mu nu)h_{\mu \nu}=\bar{h}_{\mu \nu}. In particular, in the TT gauge, the Lorenz gauge condition reduces to del_(nu)h^(nu)_(mu)=0\partial_{\nu} h^{\nu}{ }_{\mu}=0. The symmetric tensor h_(mu nu)h_{\mu \nu} has ten independent components and the TT gauge contains eight conditions, which in fact means that only two components of h_(mu nu)h_{\mu \nu} are independent. The two independent components of h_(mu nu)h_{\mu \nu} are normally denoted h_(+)h_{+}and h_(xx)h_{\times}, which correspond to the two independent polarization states of a gravitational wave.
b) First, let us consider two particles, which are influenced by a plus-polarized gravitational wave along the zz-direction such that
where h_(mu nu)h_{\mu \nu} is the given gravitational wave. We observe that this gravitational wave affects neither S_(0)S_{0} nor S_(3)S_{3}. Thus, the only effect on the geodesics is taking place in the xx - and yy-directions. Without loss of generality, we can therefore assume that z=0z=0. In this case, the gravitational wave is plus-polarized only, i.e., h_(xx)=0h_{\times}=0, so that the geodesic equation simplifies to two equations for S_(1)=-S^(1)S_{1}=-S^{1} and S_(2)=-S^(2)S_{2}=-S^{2}, namely
Next, the measured distance Delta x-=S_(1)(t)\Delta x \equiv S_{1}(t) between the two particles, which was initially the distance Deltax_(0)-=S_(1)(0)\Delta x_{0} \equiv S_{1}(0) along the xx-direction, will be
which means that the relative distance delta x-=Delta x-Deltax_(0)\delta x \equiv \Delta x-\Delta x_{0} between the two particles oscillate with ff. This does not mean that the positions of the particle coordinates change, but the coordinates themselves oscillate.
Finally, assuming dd to be the measured distance Delta x\Delta x and LL the initial distance Deltax_(0)\Delta x_{0} between the two particles, we obtain
{:(3.1962)(d)/(L)=1-(1)/(2)*h_(0)sin(2pi ft)quad=>quad d=[1-(1)/(2)h_(0)sin(2pi ft)]L:}\begin{equation*}
\frac{d}{L}=1-\frac{1}{2} \cdot h_{0} \sin (2 \pi f t) \quad \Rightarrow \quad d=\left[1-\frac{1}{2} h_{0} \sin (2 \pi f t)\right] L \tag{3.1962}
\end{equation*}
which is what we wanted to show.
2.134
First, consider the energy-momentum tensor T^(mu nu)T^{\mu \nu}. Using T^(mu nu)T^{\mu \nu} and its conservation law, i.e., grad_(nu)T^(mu nu)=0\nabla_{\nu} T^{\mu \nu}=0, we can write
Note that in the linearized approximation, grad_(v)T^(mu nu)=0\nabla_{v} T^{\mu \nu}=0 reduces to del_(nu)T^(mu nu)=0\partial_{\nu} T^{\mu \nu}=0. For the spatial components of T^(mu nu)T^{\mu \nu}, using the fact that del_(0)(T^(i0)x^(j))-(del_(0)T^(i0))x^(j)=0\partial_{0}\left(T^{i 0} x^{j}\right)-\left(\partial_{0} T^{i 0}\right) x^{j}=0, we have
where we used del_(0)(T^(i0)x^(j))=(del_(0)T^(i0))x^(j)+T^(i0)del_(0)x^(j)=(del_(0)T^(i0))x^(j)\partial_{0}\left(T^{i 0} x^{j}\right)=\left(\partial_{0} T^{i 0}\right) x^{j}+T^{i 0} \partial_{0} x^{j}=\left(\partial_{0} T^{i 0}\right) x^{j}. Similarly, for the T^(i0)x^(j)T^{i 0} x^{j} term, we use the same rewriting technique, i.e.,
Since the energy-momentum tensor is symmetric, i.e., T^(mu nu)=T^(nu mu)T^{\mu \nu}=T^{\nu \mu}, we can then combine the expressions for T^(ij)T^{i j} and widehat(T^(i0)x^(j))\widehat{T^{i 0} x^{j}}, and thus, we obtain
In addition, using that del_(0)del_(k)=del_(k)del_(0)\partial_{0} \partial_{k}=\partial_{k} \partial_{0} and del_(0)[(del_(0)T^(00))x^(i)x^(j)]=(del_(0)^(2)T^(00))x^(i)x^(j)\partial_{0}\left[\left(\partial_{0} T^{00}\right) x^{i} x^{j}\right]=\left(\partial_{0}{ }^{2} T^{00}\right) x^{i} x^{j}, we have
where MM is the mass of the binary, rr is the distance to the binary, and PP is the period of one orbit of the binary, i.e., the orbital period. In conventional units, the
amplitude of the gravitational waves can be written as [see, e.g., K.D. Kokkotas, Gravitational Waves, Acta Phys. Polon. B 38, 3891 (2007)]
where f-=1//Pf \equiv 1 / P is the frequency.
a) Using M≃2.8M_(o.)M \simeq 2.8 M_{\odot} (where M_(o.)≃1.988435*10^(30)kgM_{\odot} \simeq 1.988435 \cdot 10^{30} \mathrm{~kg} is the solar mass), r=5kpcr=5 \mathrm{kpc}, and P=1hP=1 \mathrm{~h} (which means that f=1//3600Hzf=1 / 3600 \mathrm{~Hz} ), we find that
b) Again, using M≃2.8M_(o.)M \simeq 2.8 M_{\odot}, but r=15Mpcr=15 \mathrm{Mpc} and P=0.02sP=0.02 \mathrm{~s} (which means that f=50Hzf=50 \mathrm{~Hz} ), we find that
which is basically the same result as in a).
c) For a binary system, assuming circular binary orbits, Kepler's third law provides a direct and accurate estimate for the orbital separation distance between the two binaries, i.e.,
where G≃6.674*10^(-11)Nm^(2)//kg^(2)G \simeq 6.674 \cdot 10^{-11} \mathrm{~N} \mathrm{~m}^{2} / \mathrm{kg}^{2} is Newton's gravitational constant, M_(1)M_{1} and M_(2)M_{2} are the masses of the two binaries, respectively, R_(1)R_{1} and R_(2)R_{2} are the respective distances to their common orbital center, and PP is the orbital period. Assuming M_(1)=M_(2)-=M//2M_{1}=M_{2} \equiv M / 2 and R_(1)=R_(2)-=RR_{1}=R_{2} \equiv R, i.e., the binaries are equally heavy and they are at the same distance compared to their orbital center, we obtain (half of) the orbital separation distance RR between the two binaries as
{:(3.1982)R=(1)/(2)root(3)((GMP^(2))/(4pi^(2))):}\begin{equation*}
R=\frac{1}{2} \sqrt[3]{\frac{G M P^{2}}{4 \pi^{2}}} \tag{3.1982}
\end{equation*}
Inserting M=2.8M_(o.)M=2.8 M_{\odot} and P=0.02sP=0.02 \mathrm{~s}, we find that
{:(3.1983)R≃155600m∼100km:}\begin{equation*}
R \simeq 155600 \mathrm{~m} \sim 100 \mathrm{~km} \tag{3.1983}
\end{equation*}
Thus, we can only hope to detect inspirals of compact binary systems (e.g., NS-NS, NS-BH, or BH-BH) with Earth-based interferometers such as LIGO.
d) For a spinning neutron star, the amplitude of the gravitational waves hh is approximately given by
{:(3.1984)h prop(2delta MR^(2)Omega^(2))/(r):}\begin{equation*}
h \propto \frac{2 \delta M R^{2} \Omega^{2}}{r} \tag{3.1984}
\end{equation*}
where delta M\delta M is the mass of a nonspherical deformation on the equator of the neutron star, RR is the radius of the neutron star, Omega-=2pi//P\Omega \equiv 2 \pi / P is the angular velocity expressed in the spin period PP, and rr is the distance to the neutron star. In conventional units, the amplitude of the gravitational waves can be written as [see, e.g., R. Prix, Gravitational Waves from Spinning Neutron Stars, in: W. Becker (ed.), Neutron Stars and Pulsars, Astrophysics and Space Science Library 357, 651-685, Springer (Berlin, 2009)]
{:(3.1985)h prop10^(2)(G)/(c^(4))*(delta MR^(2)f^(2))/(r)∼3*10^(-25)((delta MR^(2))/(10^(32)(kg)m^(2)))((f)/(100(Hz)))^(2)((100pc)/(r)):}\begin{equation*}
h \propto 10^{2} \frac{G}{c^{4}} \cdot \frac{\delta M R^{2} f^{2}}{r} \sim 3 \cdot 10^{-25}\left(\frac{\delta M R^{2}}{10^{32} \mathrm{~kg} \mathrm{~m}^{2}}\right)\left(\frac{f}{100 \mathrm{~Hz}}\right)^{2}\left(\frac{100 \mathrm{pc}}{r}\right) \tag{3.1985}
\end{equation*}
where GG is again Newton's gravitational constant, cc is the speed of light in vacuum, and ff is the spin frequency. Inserting delta M=10^(-6)M_(o.),R=10km,f=50Hz\delta M=10^{-6} M_{\odot}, R=10 \mathrm{~km}, f=50 \mathrm{~Hz}, and r=1kpcr=1 \mathrm{kpc}, we find that
where hh and ff are the amplitude and the frequency, respectively, of the gravitational waves, which are assumed to be monochromatic. Note that the components of the metric tensor play the role of the gravitational potential, so the derivatives of the components of the metric tensor act as the field, and therefore, F prop|h^(˙)|^(2)F \propto|\dot{h}|^{2}. A useful formula for an estimate of the energy flux on Earth due to gravitational waves is given by [see, e.g., I. Ciufolini (ed.) et al., Gravitational Waves, IoP (Bristol, 2001)]
where hh and ff are now the amplitude and the frequency, respectively, of the gravitational waves as measured on Earth. Inserting h=10^(-21)h=10^{-21} and f=200Hzf=200 \mathrm{~Hz} for GW150914, we find that
which is an enormous amount of energy flux compared to the observed energy flux in electromagnetic waves.
b) To estimate the energy flux in electromagnetic waves that is received at Earth from a full moon, we use the following assumptions: (i) The solar irradiance is about 1387W//m^(2)1387 \mathrm{~W} / \mathrm{m}^{2} at the surface of the Moon and the Earth; (ii) The reflectivity of the Moon is about 12%12 \%; and (iii) The average solid angle subtended by the Moon in the sky at Earth is about 6.4*10^(-5)6.4 \cdot 10^{-5} steradians, which is approximately the same as that subtended by the Sun, i.e., about 6.8*10^(-5)6.8 \cdot 10^{-5} steradians. Therefore, the energy flux from a full moon is given by
The assumptions made are not totally accurate, so reducing the solar irradiance to 1000W//m^(2)1000 \mathrm{~W} / \mathrm{m}^{2} and the reflectivity of the Moon to 10%10 \% lead to F_("full moon ")=6.4*10^(-3)F_{\text {full moon }}=6.4 \cdot 10^{-3}W//m^(2)\mathrm{W} / \mathrm{m}^{2}. Thus, the gravitational wave energy flux of GW150914 in a), i.e., F_(GW)∼F_{\mathrm{GW}} \sim12*10^(-3)W//m^(2)12 \cdot 10^{-3} \mathrm{~W} / \mathrm{m}^{2}, is between once or twice the electromagnetic wave energy flux of a full moon [F_(GW)∼(1,2)F_("full moon ")]\left[F_{\mathrm{GW}} \sim(1,2) F_{\text {full moon }}\right], although the estimated distance of GW150914 is about 400 Mpc and the distance from the Earth to the Moon is just 1.25*10^(-8)pc1.25 \cdot 10^{-8} \mathrm{pc}.
2.137
The metric for a linearly expanding spacetime is given by
{:(3.1991)ds^(2)=dt^(2)-H^(2)t^(2)dx^(2):}\begin{equation*}
d s^{2}=d t^{2}-H^{2} t^{2} d x^{2} \tag{3.1991}
\end{equation*}
We want to start at t=t_(0),x=0t=t_{0}, x=0 and arrive at t=t_(1),x=Lt=t_{1}, x=L without accelerating. This implies that we should move along a geodesic. Thus, we want to determine x(t)x(t). The Lagrangian is given by
{:(3.1999)int_(0)^(x(t))dx^(')=int_(t_(0))^(t_(1))(A)/(Htsqrt(betaH^(2)t^(2)+A^(2)))dt:}\begin{equation*}
\int_{0}^{x(t)} d x^{\prime}=\int_{t_{0}}^{t_{1}} \frac{A}{H t \sqrt{\beta H^{2} t^{2}+A^{2}}} d t \tag{3.1999}
\end{equation*}
The farthest one could go, x(t_(1))=Lx\left(t_{1}\right)=L would be following a lightlike geodesic, i.e., for beta=0\beta=0. Then, we obtain
{:(3.2001)L=int_(t_(0))^(t_(1))(A)/(Ht^(')sqrt(A^(2)))dt^(')=int_(t_(0))^(t_(1))(1)/(Ht^('))dt^(')=(1)/(H)ln((t_(1))/(t_(0))):}\begin{equation*}
L=\int_{t_{0}}^{t_{1}} \frac{A}{H t^{\prime} \sqrt{A^{2}}} d t^{\prime}=\int_{t_{0}}^{t_{1}} \frac{1}{H t^{\prime}} d t^{\prime}=\frac{1}{H} \ln \left(\frac{t_{1}}{t_{0}}\right) \tag{3.2001}
\end{equation*}
2.138
The cosmological redshift is due to the fact that signals interchanged between two observers will have different travel times if the size of the universe is altered. In the case of the 2-dimensional mini-universe, we assume that chi_(0) < chi_(1)\chi_{0}<\chi_{1}. (To treat the case of chi_(1) > chi_(0)\chi_{1}>\chi_{0}, simply substitute d chid \chi with -d chi-d \chi and chi_(1)-chi_(0)\chi_{1}-\chi_{0} with chi_(0)-chi_(1)\chi_{0}-\chi_{1}.) For a lightlike geodesic, we have ds=0d s=0, which gives
{:(3.2002)cdt=S(t)d chiquad=>quadint_(t_(0))^(t_(1))(cdt)/(S(t))=int_(chi_(0))^(chi_(1))d chi=chi_(1)-chi_(0):}\begin{equation*}
c d t=S(t) d \chi \quad \Rightarrow \quad \int_{t_{0}}^{t_{1}} \frac{c d t}{S(t)}=\int_{\chi_{0}}^{\chi_{1}} d \chi=\chi_{1}-\chi_{0} \tag{3.2002}
\end{equation*}
We now assume that a light signal is sent from chi_(0)\chi_{0} at time tt and another at time t+epsit+\varepsilon, they are received at chi_(1)\chi_{1} at times t^(')t^{\prime} and t^(')+epsi^(')t^{\prime}+\varepsilon^{\prime}. Assuming that epsi\varepsilon is small, the redshift is given by
{:(3.2004)int_(t)^(t^('))(cdt)/(S(t))=chi_(1)-chi_(0)=int_(t+epsi)^(t^(')+epsi^('))(cdt)/(S(t)):}\begin{equation*}
\int_{t}^{t^{\prime}} \frac{c d t}{S(t)}=\chi_{1}-\chi_{0}=\int_{t+\varepsilon}^{t^{\prime}+\varepsilon^{\prime}} \frac{c d t}{S(t)} \tag{3.2004}
\end{equation*}
Using the fact that epsi\varepsilon is small, this can be simplified to
where NN is the 4 -frequency of the light signal, is a constant. Since the 4 -frequency is a null vector, we also obtain g(N,N)=g_(tt)(N^(t))^(2)+g_(xx)(N^(x))^(2)=0quad LongrightarrowquadN^(t)=sqrt(-(g_(xx)(N^(x))^(2))/(g_(tt)))=(k)/(sqrt(-g_(tt)g_(xx)))g(N, N)=g_{t t}\left(N^{t}\right)^{2}+g_{x x}\left(N^{x}\right)^{2}=0 \quad \Longrightarrow \quad N^{t}=\sqrt{-\frac{g_{x x}\left(N^{x}\right)^{2}}{g_{t t}}}=\frac{k}{\sqrt{-g_{t t} g_{x x}}}.
The 4-velocity of a comoving observer is given by V=gammadel_(t)V=\gamma \partial_{t} and normalization results in
where Phi_(t)=del Phi//del t\Phi_{t}=\partial \Phi / \partial t, etc.
2.141
a) The geodesic equation can be computed from t^(˙)^(2)-exp(2t//t_(H))r^(˙)^(2)=0\dot{t}^{2}-\exp \left(2 t / t_{H}\right) \dot{r}^{2}=0, where t^(˙)=dt//d tau\dot{t}=d t / d \tau, i.e., dr//dt=+-exp(-t//t_(H))d r / d t= \pm \exp \left(-t / t_{H}\right) with r(t_(0))=0r\left(t_{0}\right)=0. Using dr//dt=d r / d t=+exp(-t//t_(H)) > 0+\exp \left(-t / t_{H}\right)>0, this implies that
{:(3.2021)r(t)=int_(0)^(r)dr^(')=int_(t_(0))^(t)e^(-t^(')//t_(H))dt^(')=[-t_(H)e^(-t^(')//t_(H))]_(t_(0))^(t)=-t_(H)(e^(-t//t_(H))-e^(-t_(0)//t_(H))):}\begin{equation*}
r(t)=\int_{0}^{r} d r^{\prime}=\int_{t_{0}}^{t} e^{-t^{\prime} / t_{H}} d t^{\prime}=\left[-t_{H} e^{-t^{\prime} / t_{H}}\right]_{t_{0}}^{t}=-t_{H}\left(e^{-t / t_{H}}-e^{-t_{0} / t_{H}}\right) \tag{3.2021}
\end{equation*}
b) For the line theta=pi//2\theta=\pi / 2 and varphi=0\varphi=0 at fixed universal time t=t_(0)=t=t_{0}= const., we have dt=0d t=0 and d Omega=0d \Omega=0, which imply that
{:(3.2022)ds^(2)=-e^(2t//t_(H))dr^(2)quad=>quadg_(rr)(t)=-e^(2t//t_(H)):}\begin{equation*}
d s^{2}=-e^{2 t / t_{H}} d r^{2} \quad \Rightarrow \quad g_{r r}(t)=-e^{2 t / t_{H}} \tag{3.2022}
\end{equation*}
Thus, the proper distance between the origin r=0r=0 and a point r > 0r>0 is given by
{:(3.2023)d_(p)(t","r)=int_(r^(')=0)^(r^(')=r)sqrt(-g_(rr)(t))dr^(')=int_(0)^(r)sqrt(e^(2t//t_(H)))dr^(')=e^(t//t_(H))int_(0)^(r)dr^(')=re^(t//t_(H)):}\begin{equation*}
d_{p}(t, r)=\int_{r^{\prime}=0}^{r^{\prime}=r} \sqrt{-g_{r r}(t)} d r^{\prime}=\int_{0}^{r} \sqrt{e^{2 t / t_{H}}} d r^{\prime}=e^{t / t_{H}} \int_{0}^{r} d r^{\prime}=r e^{t / t_{H}} \tag{3.2023}
\end{equation*}
c) The cosmological redshift relates the emitted and received wavelengths as (see solutions to Problems 2.138 and 2.139)
where G_(ij)G_{i j} are the components of a Riemannian metric. The proper length of a curve on the cosmological simultaneity is given by
{:(3.2027)ℓ=intsqrt(-g_(ij)x^(˙)^(i)x^(˙)^(j))ds=a(t)intsqrt(G_(ij)x^(˙)^(i)x^(˙)^(j))ds-=a(t)ℓ_(c)",":}\begin{equation*}
\ell=\int \sqrt{-g_{i j} \dot{x}^{i} \dot{x}^{j}} d s=a(t) \int \sqrt{G_{i j} \dot{x}^{i} \dot{x}^{j}} d s \equiv a(t) \ell_{c}, \tag{3.2027}
\end{equation*}
where ss is the curve parameter and ℓ_(c)\ell_{c} the comoving length of the curve. Fixing the end points of the curve and minimizing this length gives the proper distance d_(p)d_{p} between the points as d_(p)=a(t)d_(c)d_{p}=a(t) d_{c}, where the comoving distance d_(c)d_{c} is the minimal value of the integral
b) Hubble's law describes the observation that objects (read: galaxies) are moving away from the Earth at velocities proportional to their distances. The further the galaxies are, the faster they are moving away. The velocities of the galaxies are measured by their redshifts. The velocities arise from the expansion of the universe. Indeed, the proper distance between galaxies increases even though they are comoving.
2.143
a) If we use the given approximation, i.e., kk is small and rho≃rho_(Lambda)\rho \simeq \rho_{\Lambda}, we can approximate the first Friedmann equation as
We observe that the larger tt becomes, the smaller 1-Omega1-\Omega becomes, and thus, 1-Omega1-\Omega gets closer and closer to zero, which means that Omega\Omega gets closer and closer to one.
b) Today, we have a universe with a curvature that is very close to zero, which without inflation would need a lot of fine-tuning to maintain. After inflation, a spacetime is obtained, which is very close to being flat as per the solution to a), even if the initial spacetime had a lot of curvature.
2.144
a) The scaling of the energy density for an ideal fluid with equation-of-state parameter ww is given by rho=rho_(0)a^(-3(1+w))\rho=\rho_{0} a^{-3(1+w)}. The density of component ii today is given by
since w=-1w=-1 for a cosmological constant, w=0w=0 for a matter gas, and w=(1)/(3)w=\frac{1}{3} for a radiation gas. The ratio between the energy densities of matter and the cosmological constant is therefore given by
Note that this is not very long ago. A naïve estimate with H_(0)≃70km//s//Mpc≃H_{0} \simeq 70 \mathrm{~km} / \mathrm{s} / \mathrm{Mpc} \simeq2*10^(-18)s^(-1)2 \cdot 10^{-18} \mathrm{~s}^{-1} gives a couple of billion years ago (similar to the birth of the solar system).
b) Similarly, the ratio between matter and radiation energy densities is given by
{:(3.2037)(rho_(m))/(rho_(r))=(Omega_(m))/(Omega_(r))(a^(-3))/(a^(-4))=(Omega_(m))/(Omega_(r))a.:}\begin{equation*}
\frac{\rho_{m}}{\rho_{r}}=\frac{\Omega_{m}}{\Omega_{r}} \frac{a^{-3}}{a^{-4}}=\frac{\Omega_{m}}{\Omega_{r}} a . \tag{3.2037}
\end{equation*}
{:(3.2041)rho_(c)=(3H^(2))/(8pi G)quad" and "quadrho_(0i)=rho_(0c)Omega_(i)",":}\begin{equation*}
\rho_{c}=\frac{3 H^{2}}{8 \pi G} \quad \text { and } \quad \rho_{0 i}=\rho_{0 c} \Omega_{i}, \tag{3.2041}
\end{equation*}
Figure 3.18 The scale factor a(t)a(t) for the solution to Problem 2.145. The solution for Omega_(Lambda)=1\Omega_{\Lambda}=1 and Omega_(m)=0\Omega_{m}=0 is also shown for times t > t_(0)t>t_{0} for comparison.
we find that
The result of numerically evaluating this expression is shown in Figure 3.18. Note that the solution quickly becomes dominated by the cosmological constant after t=t_(0)t=t_{0} as the matter component dilutes. However, as the cosmological constant is smaller by a factor of 0.7 compared to the case where Omega_(Lambda)=1\Omega_{\Lambda}=1, the exponential factor in the large time limit (and therefore, H(t rarr oo)H(t \rightarrow \infty) ) is also smaller by a factor 0.7 .
Unlike the Omega_(Lambda)=1\Omega_{\Lambda}=1 solution, our solution has a time tt for which a(t)=0a(t)=0. Defining this time as t=0t=0, we find that (from looking at where the graph intersects zero)
Current experimental data restricts H_(0)H_{0} to be around 70km//s//Mpc70 \mathrm{~km} / \mathrm{s} / \mathrm{Mpc} or, equivalently, H_(0)≃7*10^(-11)yr^(-1)H_{0} \simeq 7 \cdot 10^{-11} \mathrm{yr}^{-1}, leading to
{:(3.2047)t_(0)≃1.3*10^(10)" years ":}\begin{equation*}
t_{0} \simeq 1.3 \cdot 10^{10} \text { years } \tag{3.2047}
\end{equation*}
estimating the universe to be around 13 billion years old.
2.146
a) The universe has a Killing vector field del_(varphi)\partial_{\varphi}. Therefore, the quantity g(del_(varphi),(gamma^(˙)))=g\left(\partial_{\varphi}, \dot{\gamma}\right)=-a^(2)varphi^(˙)=-a^{2} \dot{\varphi}= constant =-k=-k. The 4 -velocity of a comoving observer is given by U=del_(t)U=\partial_{t}, and therefore, at t=t_(0)t=t_{0}
In order for the object to complete an entire lap around the universe in finite time t_(1)t_{1}, we therefore require that
{:(3.2052)2pi=int_(0)^(2pi)d varphi=int_(t_(0))^(t_(1))(k)/(asqrt(k^(2)+a^(2)))dt:}\begin{equation*}
2 \pi=\int_{0}^{2 \pi} d \varphi=\int_{t_{0}}^{t_{1}} \frac{k}{a \sqrt{k^{2}+a^{2}}} d t \tag{3.2052}
\end{equation*}
for some finite t_(1)t_{1}. In other words, we have
b) We know that g((gamma^(˙)),del_(varphi))=-a^(2)varphi^(˙)=-kg\left(\dot{\gamma}, \partial_{\varphi}\right)=-a^{2} \dot{\varphi}=-k is constant and that t^(˙)=sqrt(1+k^(2)//a^(2))\dot{t}=\sqrt{1+k^{2} / a^{2}}. The relative velocity v(t)v(t) at time tt is therefore given by
where it has been assumed that a_(0)=1a_{0}=1 and used that sqrt((8pi G)/(3))=(H_(0)^(2))/(rho_(0,c))\sqrt{\frac{8 \pi G}{3}}=\frac{H_{0}^{2}}{\rho_{0, c}}. This is a separable differential equation and we find that
{:(3.2059)t_(0)-t_(1)=int_(t_(1))^(t_(0))dt=(1)/(H_(0))int_(a_(1))^(a_(0))(da)/(sqrt(Omega_(m)a^(-1)+Omega_(r)a^(-2)))=(1)/(H_(0))int_(a_(1))^(1)(ada)/(sqrt((1-x)a+x)):}\begin{equation*}
t_{0}-t_{1}=\int_{t_{1}}^{t_{0}} d t=\frac{1}{H_{0}} \int_{a_{1}}^{a_{0}} \frac{d a}{\sqrt{\Omega_{m} a^{-1}+\Omega_{r} a^{-2}}}=\frac{1}{H_{0}} \int_{a_{1}}^{1} \frac{a d a}{\sqrt{(1-x) a+x}} \tag{3.2059}
\end{equation*}
since we work under the assumption that a_(0)=1a_{0}=1. Performing this integral, e.g., by partial integration, leads to
The energy density ratio between matter and radiation scales as rho_(m)//rho_(r)∼a(t)\rho_{m} / \rho_{r} \sim a(t) due to rho_(m)propa^(-3)\rho_{m} \propto a^{-3} and rho_(r)propa^(-4)\rho_{r} \propto a^{-4}. Since
The time since matter-radiation equality is therefore just slightly less than the age (2)/(3H_(0))\frac{2}{3 H_{0}} of a fully matter-dominated universe.
2.148
For an expanding universe, a^(˙) > 0\dot{a}>0, which implies that a^(˙)^(3) > 0\dot{a}^{3}>0, and thus, the requirement that |Omega_(K)|\left|\Omega_{K}\right| decreases with time is given by
That |Omega_(K)|\left|\Omega_{K}\right| decreases with time is therefore equivalent to the scale factor growth accelerating. The Friedmann acceleration equation is
The condition on ww for |Omega_(K)|\left|\Omega_{K}\right| to decrease with time is therefore w < -(1)/(3)w<-\frac{1}{3}. In particular, note that this is the case for a cosmological constant, where w=-1w=-1.
2.149
a) From the evolution equations, we find that
The equation of motion for the scalar field phi\phi itself is found by varying the action with respect to phi\phi only. From the variation of the action derived in a), we then find
This is the equation of motion for phi\phi.
b) With del_(i)phi=0\partial_{i} \phi=0 and del_(t)phi=phi^(˙)\partial_{t} \phi=\dot{\phi} and using the result of Problem 2.60, we find that
where U_(mu)=delta_(mu)^(0)U_{\mu}=\delta_{\mu}^{0}. We note that g^(mu nu)U_(mu)U_(nu)=1g^{\mu \nu} U_{\mu} U_{\nu}=1 and UU is therefore a 4-velocity field. Comparing this to the stress-energy tensor of an ideal fluid T_(mu nu)=U_(mu)U_(nu)(rho_(0)+p)-T_{\mu \nu}=U_{\mu} U_{\nu}\left(\rho_{0}+p\right)-g_(mu nu)pg_{\mu \nu} p, we obtain
{:(3.2079)p=(1)/(2)*phi^(˙)^(2)-V(phi)quad" and "quadrho_(0)=(1)/(2)*phi^(˙)^(2)+V(phi):}\begin{equation*}
p=\frac{1}{2} \cdot \dot{\phi}^{2}-V(\phi) \quad \text { and } \quad \rho_{0}=\frac{1}{2} \cdot \dot{\phi}^{2}+V(\phi) \tag{3.2079}
\end{equation*}
The equation-of-state parameter is therefore given by
Note that when phi^(˙)^(2)≪V(phi)\dot{\phi}^{2} \ll V(\phi), we find w≃-1w \simeq-1. This is the equation-of-state parameter for a cosmological constant.
